IN   MEMORIAM 
FLORIAN  CAJO 


GEOMETRY: 

TEE  ELEMENTS  OF  EUCLID  AND  LEGENDRE 

SIMPLIFIED  AND  ARRANGED 
TO  EXCLUDE  FROM  GEOMETRICAL  REASONING 

WITH  THE 

ELEMEx\TS  OF  PLAXE  AND  SPHEPJCAL  TRIGONOMETRY, 

ANT) 

EXERCISES  Ii\  ELEMENTARY  GEOMETRY  AND  TRIGONOMETRY. 

ADAPTED   FOK   SCHOOLS   AND   COLLEGES. 


•  L-AWR^i^CE'  S:'  BENSON, 

n 

Author  of  "The  Truth  of  the  Bible  Upheld,"— London,  18G4;  "Geometrical  Disquisi- 
tions,"— London,  1864 ;  "  Scientific  Disquisitions  Concerning  the 
Circle  and  Ellipse,"  1862. 
Member  of  the  New  York  Association  for  the  Advancement  of  Science  and  Art ;  Hon. 
Mem.  Phi  Eappa  Society,  University  of  Georgia ;  Brothers'  Society, 
Yale  College,  etc.,  etc. 


[all  rights  reserved.] 


NEW  YORK: 

PUBLISHED  FOR  THE  AUTHOR  BY  DAVTES  &  KENT, 

No.  183  WILLLAM  STREET. 
186V. 


15f 


Rev.  Thomas  A.  Boone,  Professor  in  Carolina  Female  College,  An- 
sonville,  North  Carolina,  writes : 

"  Your  new  work  on  the  Elements  op  Geometry  (Book  First)  haa 
been  submitted  to  the  President  of  Carolina  Female  College.  He  has 
examined  it  critically,  and  indorses  it  as  an  evident  advancement  of  the 
science,  in  that  it  simplifies  and  meets  the  capacity  of  learners,  retains 
all  the  essentials  of  the  science,  and  is  equally  as  competent  for  mental 
discipline  as  the  old  Eeductio  Ad  Abmrdum." 


STBREOTTPEli  AND  ElECTROTTPEIB,- 

.-,  183  Wjjaiaar  ^treeln  Nw  Y. .     • 


CAJORI 


Entered,  according  to  Act  of  Congress,  in  the  year  1867,  by 

LAWRENCE  S.  BENSON, 

In  the  Cleric's  Office  of  the  District  Court  of  the  United  States  for  the  Southern 

District  of  New  York. 


TO 

PROFESSOR  GERARDUS  BEEKMAN  DOCHARTY,  LL.D., 

COLLEGE  OP  THE  CITY  OF  NEW  YORK. 


Sm — In  permitting  me  to  inscribe  to  you  this  Treatise  of  Elementary 
Geometry,  you  do  me  great  lionor.  Your  experience  and  success  as  a 
Teaclier  and  an  Author  will  readily  enable  you  to  give  a  full  scrutiny  to 
the  design  and  compass  of  this  volume.  Much  originality  can  not  be 
expected  in  a  subject  which  has  been,  for  more  than  two  thousand  yeare, 
enriched  by  a  great  number  of  eminent  men ;  but  in  these  days  of  practi- 
cability, a  modification  of  this  science  may  be  attempted,  as  you  have 
yourself  thought  proper  to  do,  with  a  view  of  utilizing  the  important 
principles  of  Geometry,  and  presenting  them  in  such  a  manner  that 
though  "  no  royal  road  to  Geometry"  can  be  found,  the  path  to  a  know- 
ledge of  it  may  be  rendered  so  clear  that  the  impediments  wiU  be  in  tihe 
learner  himself.  And  to  remove  much  diflBculty  in  acquiring  an  easy 
acquaintance  with  its  numerous  theorems  and  problems,  I  have  thought 
proper  to  exclude  the  inelegant  Bedrictio  ad  absurdum  from  the  methods 
of  geometrical  reasoning  which  you  have  expressed — "  a  consummation 
most  devoutly  to  he  wislied"  and  which  accomplishment,  resulting  from  my 
labors,  I  now  present  for  the  benefit  and  use  of  those  whose  education  is 
in  the  future. 

I  hav3  the  honor  to  be, 

Very  respectfully,  yours, 

Lawrence  S.  Benson. 
Vvw  York,  AprU  mh,  1867. 


TESTIMONIALS. 


The  College  op  the  City  of  New  York,  » 
Cor.  Lexington  Avenue  and  23d  Street.       1 

New  York,  January  Zd,  1867. 

I  have  had  several  interviews  with  Mr.  Lawrence  S.  Benson  on 
scientific  subjects,  and  from  his  conversation,  together  with  tlie  Essays 
which  he  has  published,  I  esteem  him  an  excellent  scholar  and  fine 
mathematician.  He  has  a  desire  to  establish  the  Elements  of  Euclid  in 
all  canes,  independently  of  the  demonstration  known  as  the  Beductio  ad 
dbsurdum,  "  a  consummation  devoutly  to  be  wished." 

Whatever  aid  or  advice  you  can  render  him  in  the  furtherance  of  this 
object  will  tend  to  the  advancement  of  true  science. 

Yours  truly, 

G.   B.   DOCHARTY. 


Rooms  op  the  New  York  Association  for  the  Advancement  ) 
of  Science  and  Art,  February  )i8th,  18G7.  I 

Extract  from  the  transactions  of  the  Association  for  the  Advancement 
of  Science  and  the  Arts : 

"  At  a  meeting  of  the  New  York  Association  held  February  25, 1867, 
a  paper  on  a  new  method  of  demonstrating  the  propositions  of  Geometry, 
denominated  the  Direct  Method,  in  place  of  the  one  now  in  use,;ijid  called 
the  Indirect  Method,  was  read  by  Lawrence  S.  Benson,  Esq.,  which 
method  the  writer  proposes  to  introduce  into  Schools  and  Academies. 

"  After  the  reading  of  the  paper,  and  the  discussion  of  its  merits,  the 
subject  was  referred  to  Professor  Fox,  Principal  of  the  Department  of 
Free  Schools  of  Cooper  Union,  and  to  Professor  Cleveland  Abbe,  for  ex- 
amination and  report.  It  was  also  moved  and  carried  that  the  Report 
when  received  be  referred  to  the  Section  on  Physical  Science  for  final 
disposition. 

"  The  Section,  after  reading  the  Report  of  Professor  Fox,  the  letter  of 
Professor  Abbe,  and  the  opinion  of  Professor  Docharty,  who  had  been 
invited  to  examine  the  work,  feel  justified  in  commending  this  work  as 
worthy  of  patronage.  Professor  Fox  in  his  Report  says :  '  The  design 
of  arranging  the  Definitions,  Axioms,  and  Propositions  of  Geometry,  so 
as  to  use  only  the  Direct  Method  of  demonstration,  is  a  good  one,  and 
when  arranged  in  the  form  of  a  neat  elementary  text-book,  will  doubtless 
do  much  good,  as  the  Direct  Method  is  much  more  easily  understood  thaa 
the  Lidirect  Method,  by  beginners.' 

"  L.  D.  Gale,  M.D., 
Oen.  Sec.  of  the  New  York  Association  for  the 
Advancement  of  Science  and  tlie  Art»." 


PREFACE. 


By  way  of  preface,  I  will  state  what  I  have  done  in  this  edition,  and 
explain  the  reason  why  I  have  done  so.  I  have  used  such  propositions 
only  which  are  required  to  substantiate  the  principal  theorems  and 
problems  by  which  the  principles  of  Geometry  have  practical  applica- 
tions in  Trigonometry,  Surveying,  Mechanics,  Engineering,  Navigation, 
and  Astronomy.  I  have  generalized  the  various  propositions  found  in 
the  school  editions  of  Geometry,  and  where  particular  cases  arise  under 
such  general  propositions,  I  have  given  the  demonstrations  for  them.  I 
have  arranged  the  propositions  to  give  the  Direct  Method  of  demonstra- 
tion in  place  of  the  Beductio  ad  ahsurdum  or  Indirect  Method. 

My  reasons  for  the  foregoing  changes  are  obvious  to  the  experienced 
mind ;  considering  the  extent  and  variety  of  modem  education,  the  time 
devoted  for  the  pupils  to  acquire  rudimental  knowledge  becomes  en- 
croached upon  in  order  to  make  them  acquainted  with  its  numerous 
modifications ;  and  for  the  pupils  to  obtain  such  knowledge  of  the  rudi- 
ments as  will  enable  them  to  see  the  practical  applications  throughout 
all  their  extent  and  variety,  which  are  very  great  in  these  days  of  ad- 
vancement and  civilization,  the  rudiments  which  were  taught  centuries 
ago  must  be  so  abbreviated  as  to  contain  the  essentials  only.  When  materials 
for  instructing  the  mind  were  scant,  there  were  no  opportunities  to  make 
close  selection  of  them ;  but  now,  when  those  materials  are  plentiful,  a 
judicious  selection  of  tlie  best  becomes  imperatively  necessary.  And 
when  geometrical  principles  have  become  extended  by  the  Algebraic 
Analysis,  and  have  been  made  practicable  by  Trigonometry,  Surveying, 
Mechanics,  Engineering,  Navigation,  and  Astronomy,  the  mere  mental 
exercises,  which  were  regarded  so  beneiicent  by  the  ancients,  are  unsuited 
for  this  practical  age,  which  is  continually  bent  on  progress,  while  the 
intellect  is  sufficiently  exercised  by  utilizing  modern  acquisitions;  for 
this  reason,  I  have  reduced  the  number  of  propositions  substantiating  the 
principles  of  Geometiy,  and  I  have  classified  them  in  such  a  manner  that 
particular  cases  are  enunciated  by  general  propositions,  a  change  which 
is  likely  to  impress  on  the  pupils  the  accuracy  of  geometrical  principles, 
as  they  will  be  shown  that  geometrical  principles  are  the  same  in  all 
cases  and  under  every  circumstance. 

Many  of  the  best  geometers  have  objected  to  the  Beductio  ad  dbsurdum 
in  Geometry,  while  all  geometers  prefer  the  Direct  Method  of  demonstra- 


VI  PREFACE. 

tion.  Any  true  proposition  is  susceptible  of  being  directly  demonstrated. 
And  without  entering  into  the  merits  or  demerits  of  the  Reductio  ad  ab' 
surdum,  I  have  exchided  it  from  geometrical  reasoning,  and  have  used 
the  Direct  Method  only,  a  change  which  agrees  with  tlie  spirit  of  the  age, 
and  fulfills  the  requirements  of  progress.  I  have  omitted  the  various 
diagrams  usually  put  among  the  definitions  of  Geometry,  because  when 
a  magnitude  is  properly  defined,  the  learner  has  a  better  conception  of 
it  from  the  definition  than  any  diagram  can  give  him ;  and  the  omission 
of  the  diagrams  will  assist  the  mental  exercise  and  cultivate  the  under- 
standing of  the  learner,  which  is  the  great  object  of  geometrical  study; 
and  if  the  learner  be  made  to  draw  the  diagrams  from  the  definitions,  he 
will  be  better  instructed  than  if  they  be  given  by  the  author.  The 
time  is  not  far  distant  when  geometrical  science  may  be  attempted  with- 
out using  diagrams  in  the  demonstrations.  The  diagrams  are  auxiliaries 
to  the  mind  in  the  ascertainment  of  truth ;  they  are  not  necessary  to  the 
existence  of  truth,  and  "  Geometry  considers  all  bodies  in  a  state  of  ab- 
straction, very  different  from  that  in  which  they  actually  exist,  and  the 
truths  it  discovers  and  demonstrates  are  pure  abstractions,  hypothetical 
truths."  Hence,  diagrams  are  like  the  pebbles  used  by  Indians  in  count- 
ing, or  other  means  of  computing  before  the  principle  of  numeration  was 
discovered ;  that  when  the  intellect  of  man  becomes  more  highly  ex- 
panded and  cultivated,  diagrams  will  be  regarded  necessary  to  the  first 
conceptions  of  geometrical  knowledge,  but  altogether  unsuited  to  a  high 
development  of  geometrical  science. 

I  am  greatly  indebted  to  Hon.  S.  S.  Randall,  City  Superintendent  of 
the  Board  of  Education  of  New  York,  for  many  valuable  suggestions  in 
the  demonstrations  and  present  arrangement  of  this  work ;  and  also  under 
many  obligations  to  Professor  Docharty,  of  the  College  of  the  City  of  New 
York ;  and  to  L.  D.  Gale,  M.D.,  General  Secretary  of  the  New  York  Asso- 
ciation for  the  Advancement  of  Science  and  Art,  Cooper  Union. 

Lawrence  S.  Benson, 

61  MoETON  Street,  City  or  New  Yobk, 
April  mh,  1867. 


THE  ELEMENTS  OF  EUCLID  AND  LEGENDRE. 


BOOK  FIRST. 
ON  THE  STRAIGHT  LINE  AND  TRIANGLE. 

DEFINITIOXS. 

1.  A  definition  is  the  precise  term  by  which  one  thing  is  dis- 
tinguished from  all  other  things. 

2.  Mathematics  is  that  science  which  treats  of  those  abstract 
quantities  known  as  numbers,  symbols,  and  magnitudes. 

3.  Geometry  is  that  branch  of  Mathematics  where  the  ex- 
tensions of  magnitudes  are  considered  without  regard  to  the 
actual  existence  of  those  magnitudes. 

4.  A  m^agnitude  has  one  or  more  of  three  dimensions,  viz., 
length,  breadth,  and  thickness. 

5.  Geometers  define  a  point,  position  without  magnitude; 
but  to  give  a  point  position,  would  entitle  it  to  the  three  dimen- 
sions of  magnitude,  whereas  a  point  in  Geometry  expresses  no 
dimension. 

6.  A  diagram,  represents  the  abstractions  of  magnitudes, 
whereby  their  dimensions  are  determined,  and  geometrical 
reasoning  conducted  without  regard  to  the  actual  properties  of 
those  magnitudes. 

'7.  A  line  expresses  length  only,  and  is  capable  of  two  con- 
ditions— it  can  be  straight  or  curved ;  when  its  length  is  always 
in  one  direction,  it  is  straight ;  but  when  there  is  a  continual 
variation  in  the  direction  of  its  length,  it  is  curved,  or  in  brevity 
called  a  curve. 

Scholiiim.  A  straight  line  can  not  be  defined  as  having  all 
its  points  in  the  same  direction,  because  the  points  of  a  line  are 
its  extremities,  and  the  extremities  of  a  curved  line  can  be 


8  THE   ELEMENTS   OP  [boOK   I. 

placed  on  a  straight  line,  and  in  this  case  the  definition  would 
not  distinguish  a  straight  line  from  a  curve.  And  if  a  line  be 
regarded  composed  of  points,  this  would  infer  that  a  point  has 
dimension  ;  but  the  intersection  of  lines  is  a  point,  which,  how- 
ever, does  not  give  position  to  the  point,  because  a  line  is  au 
abstraction,  and  position  implies  actual  existence. 

8.  A  surface  expresses  an  inclosure  by  not  less  than  three 
straight  lines,  or  by  one  curved  line,  or  by  one  straight  line  and 
one  curved  line ;  consequently  a  surface  has  breadth  and  length, 
and  the  extremities  of  surfaces  are  lines,  and  the  intersection  of 
one  surface  with  another  is  a  line. 

Scho.  A  plane  surface,  or  sometimes  called  a  plane^  is  one  in 
which  any  line  can  be  drawn  wholly  in  the  surface;  and  a 
c*rved  surface  is  one  in  which  a  curve  only  can  be  drawn 
wholly  in  the  surface  in  the  direction  of  the  curvature. 

9.  A  volume  or  solid  expresses  an  inclosure  made  by  sur- 
faces, and  has  breadth,  length,  and  thickness ;  the  extremities 
of  a  volume  are  surfaces,  and  the  intersection  of  one  volume 
with  another  is  a  surface. 

10.  An  angle  is  formed  by  two  straight  lines  meeting  each 
other ;  the  point  of  intersection  of  the  lines  is  called  the  vertex 
of  the  ano-le.  When  one  straicjht  line  meets  another  straight 
line,  so  as  to  make  two  adjacent  angles,  these  angles  are  right 
angles  when  they  are  equal ;  and  when  one  angle  is  greater  than 
the  other  angle,  the  greater  angle  is  an  obtuse  angle,  and  the 
less  angle  is  an  acute  angle.  The  straight  line  which  makes 
the  two  adjacent  angles  equal  is  the  perpendicular  to  the  other 
6trai<;ht  line. 

11.  When  two  straight  lines  on  the  same  plane  never  meet 
each  other  on  whichever  side  they  be  produced,  they  are  called 
parallel  lines. 

12.  Rectilinear  surfaces  are  contained  by  straight  lines,  and 
are  called  polygons ;  when  a  polygon  has  three  sides,  it  is  a 
triangle ;  when  it  has  four  sides,  it  is  a  quadrilateral ;  when 
it  has  five  sides,  it  is  z,  pentagon  ;  when  it  has  six  sides,  it  is  a 
hexagon  /  when  it  has  seven  sides,  it  is  a  heptagon  /  when  it 
has  eight  sides,  it  is  an  octagon  ;  when  it  has  nine  sides,  it  is  an 
enneagon  /  when  it  has  ten  sides,  it  is  a  decagon  ;  and  so  on, 
being  distinguished  by  particular  names  derived  from  the  Greek 


BOOK   I.]  EUCLID    AND    LEGENDRE.  9 

language,  denoting  the  number  of  angles  formed  by  the  sides. 
The  straight  line  drawn  through  two  remote  angles  of  a  poly- 
gon of  four  or  more  sides,  is  a  diagonal. 

13.  When  the  triangle  has  its  three  sides  equal,  it  is  equi' 
lateral;  when  two  of  its  sides  are  equal,  it  is  isosceles;  and 
when  its  sides  are  unequal,  it  is  scalene.  When  its  angles  are 
equal,  it  is  eqidangular  y  when  one  of  its  angles  is  a  right  angle, 
it  is  right-angled  ;  when  one  of  its  angles  is  an  obtuse  angle,  it 
is  obtuse-angled ;  and  when  all  its  angles  are  acute,  it  is  acute- 
angled. 

14.  When  a  quadrilateral  has  its  opposite  sides  parallel,  it  is 
2l  parallelogram  ;  when  it  has  two  sides  only  parallel,  or  none 
of  its  sides  parallel,  it  is  a  trapezium. 

15.  When  a  parallelogram  has  a  right  angle,  it  is  a  rectangle  ; 
when  it  has  two  adjacent  sides  equal,  but  no  right  angle,  it  is  a 
rhombus.  When  the  rectangle  has  its  sides  equal,  it  is  a 
square  /  and  when  its  opposite  sides  only  are  equal,  it  is  an  ob- 
long. When  the  parallelogram  has  its  opposite  sides  only 
equal,  and  no  right  angle,  it  is  a  rhomboid. 

16.  A  plane  surface  contained  by  one  line  is  a  circle  when 
every  part  of  the  line  is  equally  distant  from  a  point  in  the  sur- 
face ;  the  point  is  the  center  of  the  circle,  and  the  line  is  the 
circumference. 

17.  The  straight  line  drawn  from  the  center  to  the  circum- 
ference is  the  radius  ;  the  straight  line  drawn  from  one  part  of 
the  circumference  through  the  center  to  another  part  of  the 
circumference  is  the  diameter,  which  divides  the  circle  and  cir- 
cumference each  into  two  equal  parts.  When  the  straight  line 
does  not  pass  through  the  center,  it  is  a  chord. 

18.  That  portion  of  the  circle  contained  by  the  semicircnm- 
ference  and  diameter  is  a  semicircle;  and  that  portion  con- 
tained by  the  chord  and  a  part  of  the  circumference  is  a  seg- 
ment;  a  part  of  the  circumference  is  an  arc. 

19.  If  the  vertex  of  an  angle  be  the  center  of  a  circle,  that 
part  of  the  circumference  intercepted  by  the  sides  of  the  angle 
will  give  the  value  of  the  angle ;  hence,  the  angle  is  measured 
by  an  arc  when  its  vertex  is  the  center  of  the  circle,  liut 
when  the  vertex  is  in  the  circumference,  the  angle  is  subtended 
by  the  arc  intercepted  by  its  sides ;  hence,  equal  angles  will  be 


10  THE    ELEMENTS   OF  [BOOK   I. 

measured  by  equal  arcs,  and  subtended  by  equal  arcs ;  therefore 
equal  arcs  measure  or  subtend  equal  angles. 

20.  Two  arcs  are  supplementary  when  both  together  are 
equivalent  to  the  semicircumferenoe.  And  two  angles  are  sup- 
plementary when  both  together  are  equivalent  to  two  right 
angles,  and  complementary  when  equivalent  to  one  right 
angle. 

21.  Things  are  equal  when  they  have  equal  magnitudes  and 
when  they  coincide  in  all  respects;  and  are  equivalent  when 
they  have  equal  magnitudes,  but  do  not  coincide  in  all  respects. 

22.  The  term,  each  to  each,  or  sometimes  respectively,  is  a 
limiting  expression,  and  is  used  to  denote  the  equality  of  lines 
or  magnitudes  taken  in  the  same  order;  for  without  this  quali- 
fication, two  lines  or  magnitudes  said  to  be  equal  to  two  other 
lines  or  magnitudes,  would  imply  that  their  sums  are  equal, 
when  it  would  be  desirous  of  meaning  that  they  are  equal  in 
the  same  order  in  which  they  are  expressed — a  difference  very 
important  in  the  demonstration  of  a  proposition. 

23.  A  proposition  is  demonstrated  by  superpositio7i  when 
one  figure  is  supposed  applied  to  another,  which  is  done  in  the 
first  case  of  the  third  proposition  of  this  book. 

24.  One  proposition  is  the  converse  of  another  when,  in  the 
language  of  logic,  the  subject  of  the  latter  is  the  predicate  of 
the  former,  and  the  predicate  of  the  latter  is  the  subject  of  the 
former. 

METHOD    OF   REASONING. 

1.  From  the  foregoing  definitions,  it  is  shown  that  the 
straight  line  and  curve  have  certain  relations,  uses,  and  prop- 
erties which  are  important  to  be  known.  And  in  order  that 
these  relations,  uses,  and  properties  may  be  satisfactorily  inter- 
preted, there  are  certain  terms,  expressive  of  certain  facts  or 
states  of  knowledge,  by  means  of  which  the  mind  intuitively 
perceives  a  connection  between  the  things  known  and  those  for 
elucidation,  such  as  axioms^  hypotheses^  and  postulates/  as 
demonstrations,  theorems, problems,  and  lemmas,'  as  corollaries 
and  scholiums.  With  the  assistance  of  these,  the  mind  is 
carried  step  by  step  in  all  its  investigation  of  extension,  and  is 
able  to  discover  by  such  investigation  the  properties,  uses,  and 
relations  of  geometrical  magnitudes.     They  are  the  data  by 


BOOK  I.]  EUCLID  AND  LEGENDRE.  H 

which  the  hidden  truths  are  revealed.  Upon  them  a  system  of 
logic  or  argumentation  is  conducted,  and  by  the  conformity  of 
the  arguments  and  conclusions  with  the  accepted  truths,  we 
have  the  science  of  Geometry. 

2.  Proposition  in  Geometry  is  a  general  term,  expressing  the 
subjects  to  be  considered,  and  is  either  a  problem  or  theorem. 
When  it  is  the  first,  there  is  something  required  to  be  per- 
formed, such  as  drawing  a  line  or  constructing  a  figure ;  and 
whatever  points,  lines,  angles,  or  other  magnitudes  are  given  to 
efiect  the  purpose,  they  are  the  data  of  the  problem  ;  and  when 
it  is  the  latter,  a  truth  is  proposed  for  demonstration,  and 
whatever  is  assumed  or  admitted  to  be  true,  and  from  which 
the  proof  is  to  be  derived,  is  the  hypothesis. 

3.  Demonstration  consists  in  evident  deductions  from  clear 
premises,  whereby  the  conclusion  corroborates  the  premises  and 
shows  the  argumentativeness  of  the  deductions.  In  the  course 
of  demonstration,  reference  is  often  made  to  some  previous 
proposition  or  definition. 

4.  Sometimes  inferences  arise  involving  another  principle, 
but  do  not  require  any  long  process  of  reasoning  to  establish 
their  truth — these  are  corollaries.  Any  remark  made  from  the 
demonstration  of  a  proposition  is  a  scholium.  A  proposition 
which  is  preparatory  to  one  or  more  propositions,  and  is  of  no 
other  use,  is  a  lemma. 

5.  And  for  the  establishment  of  a  proposition,  there  are  four 
things  required,  viz. :  the  general  enunciation,  the  particular 
enunciation,  the  construction,  and  the  demonstration. 

6.  The  hypotheses  of  demonstration  are  known  as  axiom,  and 
postulate;  the  former  is  assumed  to  prove  the  truth  of  a  theorem, 
and  the  latter  is  granted  to  pei'form  the  requisites  of  a 
problem. 

7.  An  axiom  is  so  evidently  clear,  that  no  process  of  reason- 
ing can  make  it  more  clear ;  its  truth  is  so  easily  recognized  by 
the  human  mind,  that  so  soon  as  the  terms  by  which  it  is  ex- 
pressed are  understood,  it  is  admitted;  for  instance,  it  is  as- 
sumed as 

AXIOMS. 

1.  Things  which  are  equal  to  the  same,  or  to  equals,  are  equal 
to  one  another. 


12  THE   ELEMENTS   OF  [bOOK  I. 

2.  If  equals  or  the  same  be  added  to  equals,  the  wholes  are 
equal, 

3.  If  equals  or  the  same  be  taken  from  equals,  the  remainders 
are  equal. 

4.  If  equals  or  the  same  be  added  to  unequals,  the  wholes 
are  unequal. 

5.  If  equals  or  the  same  be  taken  from  unequals,  the  re- 
mainders are  unequal. 

6.  Things  which  are  doubles  of  the  same,  or  of  equals,  are 
equal  to  one  another. 

I.  Things  which  are  halves  of  the  same,  or  of  equals,  are 
equal  to  one  another. 

8.  Magnitudes  which  exactly  coincide  with  one  another  are 
equal. 

9.  The  whole  is  greater  than  its  part. 

10.  The  whole  is  equal  to  all  its  parts  taken  together. 

II.  All  right  angles  are  equal  to  one  another. 

12.  If  a  straight  line  meet  two  other  straight  lines  which  are 
in  the  same  plane,  so  as  to  make  the  two  interior  angles  on  the 
same  side  of  it,  taken  together,  less  than  two  right  angles,  these 
straight  lines  shall  at  length  meet  upon  that  side,  if  they  be 
continually  produced. 

These  are  the  self-evident  truths  used  by  Euclid  for  geomet- 
rical demonstration  ;  but  if  the  first  eleven  be  considered  for 
awhile,  it  will  be  seen  that  they  can  be  reduced  to  two  general 
axioms,  viz.,  things  which  are  equal  to  the  same  are  equal,  and 
things  which  are  not  equal  to  the  same  are  unequal ;  because 
when  we  add,  subtract,  multiply,  or  divide  equals,  the  equality 
in  each  case  is  not  destroyed;  hence  in  each  case  equal  to 
one  another.  And  when  we  add  unequals  to  or  subtract  un- 
equals from  equals,  the  sums  or  remainders  are  not  equal  to  the 
same,  hence  unequal  to  one  another.  And  magnitudes  which 
exactly  coincide  with  one  another  are  equal  to  the  same,  hence 
equal ;  a  whole  and  a  part  are  not  equal  to  the  same,  hence  are 
unequal ;  Avhile  a  whole  and  all  its  parts  are  equal  to  the  same, 
hence  are  equal.  From  the  definition  of  right  angles,  it  is 
seen  that  when  a  straight  line  meets  another  straight  line,  so  as 
to  make  the  two  adjacent  angles  formed  by  them  equal  to  one 


BOOK   I.]  EUCLID   AND   LEGENDRE.  13 

another,  the  two  adjacent  angles  are  right  angles;  then  these 
two  right  angles  are  equal ;  and  since  all  right  angles  agree  with 
the  definition,  they  are  equal  to  the  same  thing,  hence  equal  to 
one  another.  But  the  twelfth  axiom  is  not  self-evident,  be- 
cause the  converse  has  been  demonstrated,  viz.:  that  two 
straisrht  lines  which  meet  one  another  make  with  any  third  line 
the  interior  angles  less  than  two  right  angles.  Geometers  perceiv- 
ing this  blemish  in  the  Elements  of  Euclid,  have  endeavored  in 
many  ways  to  remove  it,  but  without  complete  success.  They 
employed  three  methods  for  this  purpose :  1,  By  adopting  a 
new  definition  of  parallel  lines.  2.  By  introducing  a  new 
axiom.  3.  By  reasoning  from  the  definition  of  parallel  lines, 
and  the  properties  of  lines  already  demonstrated.*  The  diffi- 
culty with  parallel  lines  is,  that  geometers  have  confounded  a 
definition  with  a  proposition.  Definition  11  is  perfectly  legiti- 
mate, as  it  simply  defines  what  kind  of  lines  are  parallel ;  but 
when  it  is  inferred  from  it  that  these  lines  are  equally  distant 
from  each  other,  this  is  no  axiomic  inference,  because  the  curve 
and  its  asymptote  are  two  lines  which  never  meet,  however  far 
they  be  produced  on  the  same  plane,  but  they  are  not  equally 
distant  from  each  other ;  hence  the  inference  that  parallel  lines 
are  equally  distant,  embodies  a  question  which  requires  a  dem- 
onstration to  establish ;  and  to  establish  this  question  has  given 
perplexity  to  geometers,  for  though  they  have  proven  the  lines 
equally  distant  at  particular  points,  they  have  not  proven  them 
so  at  every  point;  and  here  consists  the  incompleteness  of  their 
demonstrations,  and  here  is  required  some  general  demonstra- 
tion which  will  embrace  every  part  of  the  lines,  however  so  far 
they  be  produced  on  the  same  plane.f 

8.  A  postulate  is  a  problem  so  easy  to  perform  that  it  does 
not  require  any  explanation  of  the  manner  of  doing  it,  so  that 
the  geometer  reasonably  expects  the  method  to  be  known ;  for 
instance,  it  is  granted  as — 

*  See  notes  to  Playfair's  Euclid,  Legendre's  Geometry,  Leslie's  Geom- 
etry, the  ea-cursus  to  the  tirst  book  of  Camerer's  Euclid,  Berlin,  1825 ; 
Col.  P.  Thomson's  Oeometry  xoithout  Axioim,  Professors  Thomson's  and 
Simson's  editions  of  Euclid — London,  Glasgow,  and  Belfast. 

f  See  fifteenth  and  nineteenth  propositions  cf  this  book. 


14  THE   ELEMENTS   OF  [bOOK   I. 

POSTULATES, 

1.  That  a  straight  line  can  be  drawn  from  any  one  point  to 
any  other  point. 

2.  That  a  terminated  straight  line  may  be  extended  to  any 
length  in  a  straight  line. 

3.  That  a  circle  may  be  described  from  any  center,  at  any 
distance  from  that  center. 

EXPLAIS'ATION    OF    SIGSTS. 

In  Algebra,  the  sign  +, called  Plus  {more  Jy), placed  between 
the  names  of  two  magnitudes,  is  used  to  denote  that  these  mag- 
nitudes are  added  together ;  and  the  sign  — ,  called  Minus 
{less  hy)^  placed  between  them,  to  signify  that  the  latter  is  taken 
from  the  former.  The  sign  z=,  which  is  read  equal  to,  signifies 
that  the  quantities  between  which  it  stands  are  equal  to  one 
another.  The  sign  =o=,  signifies  that  the  quantities  between 
which    it  stands   are  equivalent  to  one  another. 

In  the  references,  the  Roman  numerals  denote  the  book,  and 
the  others,  when  no  word  is  annexed  to  them,  indicate  the 
proposition ;  otherwise  the  latter  denote  a  definition,  postulate, 
or  axiom,  as  specified.  Thus,  III.  16  means  the  sixteenth 
proposition  of  the  third  book ;  and  I.  ax.  2,  the  second  axiom 
of  the  first  book.  So  also  hyp.  denotes  hypothesis,  and  const, 
construction. 


PROPOSITIONS. 


Prop.  I. — Problem. — To  describe  an  isosceles  triangle  on  a 
finite  straight  line  given  in  position. 

Let  AX  be  the  given  straight  line;  it  is  required  to  describe 
an  isosceles  triangle  having  its  base  on  AX. 

From  a  point  C,  without  the  line  AX  as 
a  center,  and  a  radius  CA  (I.  post.  3),  de- 
scribe a  circle  ABED,  cutting  the  line  AX 
in  two  points  A  and  B;  draw  from  these 
points  the  straight  lines  AE  and  BD  (I. 
post.  1)  passing  through  the  center  of  the 
circle ;  the  triangle  ACB  is  the  one  required. 
Because  C  is  the  center  of  the  circle  ABED  (I.  def  16),  CA 


BOOK   I.J  EUCLID    AND    LEGENDRE.  15 

is  equal  to  CB,  therefore  the  triangle  ACB  has  two  sides  equal ; 
hence  (I.  def.  13)  it  is  isosceles,  and  is  described  on  AX,  which 
was  required  to  be  done. 

Corollary  1.  But  the  angle  EAB  is  subtended  by  the  arc 
EB  (I.  def.  19),  and  the  angle  DBA  is  subtended  by  the  arc 
DA ;  since  AE  and  DB  pass  through  the  center  of  the  circle 
(const,),  they  are  both  diameters  of  the  circle  (I.  def  17) ;  hence 
the  arcs  DEB  and  ADE  are  each  a  semicii-cumference,  and  (I. 
ax.  1)  are  equal ;  therefore  the  sum  of  tlie  arcs  BE  and  ED 
is  equivalent  to  the  sum  of  the  arcs  ED  and  DA ;  the  arc  ED 
is  common  ;  hence  (I.  ax.  3)  we  have  the  arc  EB  equal  to  the 
arc  DA ;  therefore  the  angles  EAB  and  DBA  are  subtended 
by  equal  arcs,  consequently  (I.  def  19)  the  angles  are  equal. 
Hence,  in  an  isosceles  triangle,  the  angles  opposite  the  equal 
sides  are  equal. 

Cor.  2.  The  line  AE,  which  forms  with  AB  the  angle  EAB, 
intercepts  the  line  DB  at  C,  which  forms  with  AB  the  angle 
DBA,  and  the  line  DB  intercepts  AE  at  C  also.  C  being  the 
center  of  the  circle  ABED,  CB,  that  portion  of  BD  intercepted 
hy  AE,  is  equal  to  CA,  that  portion  of  AE  intercepted  by  BD 
(I.  def  16) ;  but  CB  and  CA  are  the  sides  of  the  triangle  ACB 
(I.  1)  ;  hence,  when  two  angles  of  a  triangle  are  equal,  the 
opposite  sides  to  them  are  also  equal,  and  the  triangle  is  isos- 
celes (I.  def  13). 

Pkop.  II. — Problem. —  To  describe  an  equilateral  triangle 
on  a  finite  straight  line  given  in  magnitude. 

Let  AB  be  the  given  straight  line ;  it  is  required  to  describe 
an  equilateral  triangle  having  AB  for  its  base. 

From  A  as  a  center,  and  a  radius 
AB  (I.  post.  3),  describe  the  circle 
FCD ;  and  from  B  as  a  center,  and 
a  radius  BA,  describe  the  circle 
HCE.  The  circles  having  equal 
radii  (I.  ax.  1)  are  equal ;  draw  from 
C  through  the  center  B,  CH ;  and 
from  C  through  the  center  A,  CF  (I.  post  1) ;  the  triangle  ACB 
is  the  one  required. 

Because  the  circles  have  equal  radii  (const.),  AC  is  equal  to 


16  THE   ELEMENTS    OF  [bOOK   I. 

AB,  and  CB  is  equal  to  AB ;  hence  (I.  ax.  1)  the  three  sides 
of  the  triangle  ACB  are  equal;  the  triangle  (I.  def.  13)  is  equi- 
lateral, and  is  described  on  AB,  which  was  required  to  be  done. 
Corollary  1.  If  AB  be  produced  both  ways  (I.  post.  2)  to  D 
and  E,  the  angle  DBC  is  subtended  by  the  arc  CD,  and  the 
angle  FCB  is  subtended  by  the  arc  FB;  the  arcs  OB  and  BF 
are  together  equivalent  to  the  arcs  BC  and  CD  (I.  ax.  1)  ; 
hence  (I.  1,  cor.  1)  the  arc  BF  is  equal  to  the  arc  CD,  therefore 
(I.  def  19)  the  angle  FCB  is  equal  to  the  angle  DCB.  Again: 
the  angle  ACH  is  subtended  by  the  arc  AH,  and  the  angle 
CAE  is  subtended  by  the  arc  CE.  But  (I.  ax.  1)  the  arcs  AC 
and  CE  are  equivalent  to  the  arcs  CA  and  AH,  and  (I.  ax.  3) 
the  arcs  CE  and  AH  are  equal;  therefore  (I.  def  19)  the  angles 
ACH  and  CAE  are  equal,  but  the  angle  ACH  is  the  same  as 
the  angle  FCB  ;  hence  (I.  ax.  1)  the  three  angles  of  the  triangle 
are  equal ;  therefore  in  an  equilateral  triangle  the  angles  are 
equal.  And  in  a  manner  similar  to  Cor.  2  of  the  first  proi)osi- 
tion,it  can  be  shown,  conversely^  that  when  a  triangle  has  three 
equal  angles,  the  sides  opposite  them  are  also  equal ;  hence  an 
equilateral  triangle  is  also  equiangular,  and,  conversely^  an  equi- 
angular triangle  is  also  equilateral. 

Prop.  IH. — Theorem. — If  two  triangles  have  txoo  sides  of 
the  one  equal  to  two  sides  of  the  other,  each  to  each,  and  have 
also  an  angle  in  one  equal  to  an  angle  in  the  other  simiiiarly 
situated  with  respect  to  those  sides,  the  triangles  have  their 
bases  or  remaining  sides  equal /  their  other  angles  equal,  each 
to  each,  viz.,  those  to  which  the  equal  sides  a.e  opposite,  and 
the  triangles  are  equal. 

This  general  proposition  has  four  cases,  viz. :  first,  when  the 
equal  angles  are  contained  by  the  respectively  equal  sides; 
Becond,  when  the  equal  angles  are  opposite  to  one  pair  of  the 
respectively  equal  sides ;  third,  when  the  equal  angles  are  op- 
posite to  the  other  pair  of  the  respectively  equal  sides ;  and 
fourth,  the  limitation  that  when  the  least  sides  respectively  of 
the  triangles  be  equal,  and  the  angles  opposite  the  least  sides 
"be  equal,  the  angles  opposite  the  greater  of  the  respectively 
equal  sides  must  be  of  the  same  kind,  either  both  acute,  or  not 
acute. 


BOOK  I.] 


EUCLID  AND  LEGENDRE. 


17 


First  case.  Let  ABC  and  DEF  be  the  two  triangles  having 
any  two  sides  equal,  each  to  each, 
viz.,  AC  and  CB  equal  to  EF  and 
DF,  and  the  contained  angles  ACB 
and  EFD  equal ;  the  remaining  sides 
AB  and  DE  are  equal,  tlie  angle  CBA 
opposite  AC  equal  to  the  angle  FDE 
opposite  FE,  the  angle  CAB  opposite 
CB  equal  to  the  angle  FED  opposite 
DF,  and  tlie  triangles  ABC  and  DEF 
are  equal. 

If  the  triangle  ABC  be  placed  on  the  triangle  DEF  so  that 
the  vertex  of  the  angle  ACB  will  fall  on  the  vertex  of  the 
angle  DFE,  the  angle  ACB  being  equal  to  the  angle  DFE 
(hyp.),  the  side  CB  will  fall  on  FD,  and  the  side  CA  will  fall  on 
FE ;  CB  and  FD  being  equal  (hyp.),  the  extremity  B  will  fall 
on  the  extremity  D.  CA  and  FE  being  equal  (hyp.),  the  ex- 
tremity A  will  fall  on  the  extremity  E ;  and  since  AB  is  a 
straight  line,  it  will  coincide  with  DE  (I.  def.  V),  a  straight  line 
drawn  from  D  to  E.  Therefore  the  triangle  ABC  has  its  three 
sides  coinciding  with  the  three  sides  of  the  triangle  DEF ;; 
hence  the  angle  CAB  will  fall  on  the  angle  FED,  and  be  equal 
to  it ;  the  angle  CBA  will  fall  on  the  angle  FDE,  and  be  equal 
to  it ;  consequently  the  two  triangles  have  their  three  sides  and. 
three  angles  equal,  each  to  each,  and  (I.  ax.  8)  are  equal. 

Second  case.  When  the  triangles  ABC  and  DEF  have  the 
sides  CA  and  CB  respectively  equal  to  FE  and  FD,  and  the 
angles  ABC  and  EDF  equal,  respectively  opposite  to  CA  and 
FE,  the  remaining  sides  are  equal ;  the  angle  CAB  opposite  CB 
is  equal  to  the  angle  FED  opposite  to  FD,  the  angle  ACB  op- 
posite to  AB  is  equal  to  the  angle  EFD  opposite  to  DE,  and 
the  triangles  are  equal. 

Let  the  side  DE  b 

be  put  on  AB  so 

that  D  will  fall 

on    B,    and    the 

eqnal  angles  ABC 

and  EDF  will  be 

on  different  sides  of  AB ;  join  CF  (I.  post.  1 ).    BC  and  BF  (hyp.)  are 
2 


18  THE   ELEMENTS    OF  [bOOK    I. 

equal, the  triangle  CBF  (I.  def.  13)  is  isosceles,  and  (I.  1,  cor.  l) 
the  angle  BCF  eqiial  to  the  angle  BFC;  and  hecause  CA  and, 
AF  are  equal  (hyp.),  the  triangle  CAF  (I.  def.  13)  is  isosceles, 
and  (1, 1,  cor.  1)  the  angle  ACF  is  equal  to  the  angle  AFC  ; 
then  (I.  ax.  2)  the  angles  BCF  and  ACF  are  equal  to  the  angles 
BFC  and  AFC,  or  the  angle  BCA  equal  to  the  angle  BFA 
(I.  ax.  1  and  ax.  10).  Hence  we  have  in  the  triangles  ABC 
and  ABF,  two  sides,  and  the  contained  angle  in  each  equal, 
.-«ach  to  each,  therefore  hj first  case  the  triangles  can  be  shown 
.-.equal  in  all  respects. 

Third  case.    It  can  be  proven  in  a  similar  manner  as  the  second 

*  case.     When  the  angles  CAB  and  FED  of  the  second  case  are 

,  obtuse,  and  the  angles  CBA  and  FDE  of  the  third  case  are 

.  obtuse,  the  proofs  are  given  by  the  third  axiom  of  the  first  book. 

Fourth   case.     When   the   triangles   ABC   and   DEF   have 

their  least  sides  in  each  equal — viz.,  AB  to  DE — and  another 

side  in  each,  equal,  the  angles  ACB  and  EFD  being  equal,  the 

angles  opposite  the  second  pair  of  equal  sides  must  be  both 

acute  or  both  not  acute  ;  otherwise,  two  triangles  can  be  formed 

having  two  sides  .and  an  augl'-^  in  each  equal,  each  to  each,  and 

?  the  triangles  unequal.     For,  in  the 

triano-lcs  ABC  and  ACD,  the  side 
AC    is   common,   the   angle  BAC 
equal    to    angle  CAD,  the  sides  BC 
and  CD  can  be  equal,  and  the  tri- 
angles (I.  ax.  9)  unequal;  hence  in  two  triangles  when  the 
greatest  and  least  sides  are  respectively  equal,  and  the  equal 
angles  opposite  to  the  least  sides  be  given,  the  angles  opposite 
the  greatest  sides  must  both  be  not  acute  to  determine  the  tri- 
angles ;  but  when  in  two  triangles  the  two  less  sides  of  each 
are  respectively  equal,  and  the  equal  angles  opposite  tlie  least 
sides  be  given,  the  angles  opposite  the  other  equal  sides  must 
both  be  acute,  to  determine  the  triangles. 

The  equality  of  the  triangles  can  be  proven  by  {ho  second 
and  third  cases,  using  the  second  axiom  when  the  angles  are 
acute,  and  the  third  axiom  when  the  angles  are  obtuse ;  but 
when  the  angles  are  right-angled,  the  equality  of  the  triangles 
is  shown  from  tlie  first  corollary  to  the  first  proposition  without 
those  axioms. 


!BOOK   I.] 


EUCLID   AND   LEGENDKE. 


19 


B 


Peop.  IV. — Theok. — If  the  three,  sides  of  one  triangle  be 
■equal  to  the  three  sides  of  another,  each  to  each:  (1)  the  ajigles 
■of  one  triayigle  are  equal  to  the  angles  of  the  other,  each  to  each^ 
viz.,  those  to  which  tJie  equal  sides  are  02)posite,  and  (2)  the 
triangles  are  equal. 

Let  ABC  and  DEF  be  the  two  tri-  c     p 

angles  having  then-  three  sides  equal, 
viz.,  AB  to  DE,  CA  to  FE,  and  CB 
to  FD,  the  angles  are  equal,  viz.,  ACB 
to  EFD,  CAB  to  FED,  and  CBA 
to  FDE  ;  and  the  triangles  are  equal. 

If  the   side  DE  be  placed  on  the 
side   AB   so   that   the   triangles  Avill 
fall  on  different  sides  of  AB,  D  will 
fall  on  B,  and  E  on  A,  because  DE  is  equal  to  AB,  and  the 
triangle  DEF  will 
take  the  position 
BFA,   BF    being 
the  same   as  DF, 
and  FA  the  same 
as  FE.     Join  CF,  a 

and  because  (hyp.)  BC  is  equal  to  BF,  the  angles  BCF  and 
BFC  are  equal  (I.  1,  cor.  1).  It  would  be  shown  in  a  similar 
manner  that  the  angles  FCA  and  CFA  are  equal,  therefore  (I. 
ax.  2)  the  angles  BCA  and  BFA  are  equal — that  is  (I.  ax.  1),  the 
angles  BCA  and  DFE  are  equal.  But  (hyp.)  the  sides  CB  and 
FD  are  equal,  and  the  sides  CA  and  FE,  and  it  has  been  shown 
that  the  contained  angles  are  equal,  therefore  (I.  3,  first  case) 
the  other  angles  are  equal — that  is,  CAB  to  FED  and  CBA  to 
FDE,  and  the  triangles  are  equal.  Wherefore,  if  the  three 
sides,  etc. 

Prop.  V. — Prob. —  To  bisect  a  given  angle,  that  is,  to  divide 
it  into  tico  equal  angles. 

Let  BAC  be  the  given  angle ;  it  is  required  to  bisect  it. 

From  A  as  a  center,  and  AD  less  than  AB  (I.  post.  3),  de- 
scribe the  arc  DE  ;  draw  the  chord  DE,  then  upon  DE,  on  the 
side  remote  from  A,  describe  an  equilateral  triangle  (I.  2),  DFE, 
then  join  AF ;  AF  bisects  the  angle  BAC. 


20 


THE   ELEMENTS   OF 


[book  I. 


Because  AD  is  equal  to  AE  (L  def.  16), 
and  AF  is  comraoii  to  the  two  triangles 
DAF  and  EAF,  the  two  sides  DF  and 
EF  are  equal  (I.  2) ;  therefore  the  two  tri- 
angles DAF  and  EAF  have  their  three 
sides  equal,  each  to  each,  and  the  triangles 
are  equal  (I.  4) ;  consequently  the  angle 
DAF  opposite  DF  is  equal  to  the  angle 
EAF  opposite  EF,  and  the  angle  BAG 
is  bisected  by  the  line  AF;  which  was  to  be 
done. 

OTHERWISE, 

Let  BAG  be  the  given  angle ;  in  AB  take  any  two  points  as 
B  and  D,  and  cut  off  AG  and  AE  respectively  equal  to  AB  and 
AD,  join  BE  and  GD,  and  the  straight  line  joining  the  inter- 
section of  BE  and  CD  with  the  vertex  A  bisects  BAG.  The 
proof  is  easy,  and  is  omitted  to  exercise  the  ingenuity  of  the 
pupU. 

Prop.  VI. — Prob. — To  bisect  a  given  finite  straight  line. 
Let  AB  be  the  given  line  ;  it  is  required  to  bisect  it. 
Describe  (L  2)  upon  it  an  equilateral  triangle  ABG,  and 
C  bisect  (I.  5)  the  angle  AGB  by  the  straight 

line  GD ;  AB  is  bisected  in  the  point  D. 

Because  AG  is  equal  to  GB,  and  CD 
common  to  the  two  triangles  AGD,  BGD, 
the  two  sides  AG,  GD  are  equal  to  BG, 
B  CD,  each  to  each ;  and  the  angle  AGD  is 
equal  (const.)  to  the  angle  BGD;  therefore  the  base  AD  is 
equal  (L  3)  to  the  base  DB,  and  the  line  AB  is  bisected  in  the 
point  D ;  which  was  to  be  done. 

Sc/io.  In  practice,  the  construction  is  effected  more  easily  by 
describing  arcs  on  both  sides  of  AB,  from  A  as  a  center,  and 
"with  any  radius  greater  than  the  half  of  AB ;  and  then,  by  de- 
gcribing  arcs  intersecting  them,  with  an  equal  radius,  from  B  as 
center,  the  line  joining  the  two  points  of  intersection  will  bi- 
sect AB.     The  proof  is  easy. 


Peop.  VII. — Peob. — To  draw  a  straight  litie  perpendieuiar 


BOOK   I.] 


EUCLID   AND   LKGENDRE. 


21 


to  a  given  straight  line,  from  a  giveyi  point  in  that  straight 
line. 

Let  AB  be  the  given  straight  line,  and  C  a  point  given  in  it ; 
it  is  required  to  draw  a  perpendicular  from  the  point  C. 

From  C  as  a  center,  and  a  radius 
CE  (I.  post  3),  describe  the  semicircle 
EHF;  then  (I.  def.  16)  EC  is  equal  to 
CF,  and  on  EF  (I.  2)  describe  the  ^ 
equilateral  triangle  EDF ;  then  a  line 
from  C  to  the  vez'tex  D  is  the  perpen- 
dicular required. 

Because  EC  is  equal  to  CF  (I.  def. 
16),  ED  is  equal  to  DF  (I.  2),  and  the  angle  DEC  equal  to 
angle  DFC  (I.  2,  cor.  1) ;  hence  (I.  3)  the  triangles  ECD  and 
FCD  are  equal ;  but  the  angle  ECD  is  equal  to  the  angle  FCD, 
and  therefore  (I,  def  1 0)  DC  is  perpendicular  to  AB  from  C. 


Pkop.  Vin. — Prob. — To  draw  a  straight  line  perpendicular 
to  a  given  straight  line  of  an  unlimited  length,  from  a  given 
•point  without  it. 

Let  AB  be  the  given  straight  line,  which  may  be  produced 
any  length  both  ways,  and  let  C  be  a  point  without  it.  It  is 
required  to  draw  a  straight  line  from  C 
perpendicular  to  AB. 

Take  any  point  D  upon  the  other  side 
of  AB,  and  from  the  center  C,  at  the  dis- 
tance CD,  describe  (L  post.  3)  the  circle 
ADB  meeting  AB  in  A  and  B ;  bisect  (I.  6)  AB  in  G,  and  join 
CG ;  the  straight  line  CG  is  the  perpendicular  required. 

Join  CA,  CB.  Then,  because  AG  is  equal  to  GB,  and  CG 
common  to  the  triangles  AGC,  BGC,  the  two  sides  AG,  GC 
are  equal  to  the  two,  BG,  GC,  each  to  each ;  and  the  base  CA 
is  equal  (I.  def  1 6)  to  the  base  CB ;  therefore  the  angle  CGA 
is  equal  (I.  4)  to  the  angle  CGB ;  and  they  are  adjacent  angles ; 
therefore  CG  is  perpendicular  (I.  def  10)  to  AB.  Hence,  from 
the  given  point  C  a  perpendicular  CG  has  been  drawn  to  the 
•given  line  AB ;  which  was  to  be  done. 

Scho.  This  proposition  and  the  preceding  contain  the  only 
.two  distinct  cases  of  drawing  a  perpendicular  to  a  given  straight 


22  THE   ELEMENTS   OF  [BOOK   B- 

line  through  a  given  point ;  the  first,  when  the  point  is  in  the 
line ;  the  second,  when  it  is  zcithout  it. 

In  practice,  the  construction  will  be  made  rather  more  simple 
by  describing  from  A  and  B,  when  found,  arcs  on  the  remote 
side  of  AB  from  C,  with  any  radius  greater  than  the  half  of 
AB,  and  joining  their  point  of  intersection  with  C. 

Prop.  IX. — T^eor. —  "When  one  straight  line  meets  another 
straight  line  ayid  forms  two  unequal  angles  on  the  same  side 
of  that  line,  the  two  angles  will  be  equivalent  to  two  right 
angles. 

^  Let    the   straight    line  DC    meet    the 

straight  line  AB  and  form  the  two  un- 
equal angles  DCA  and  DCB  on  the  same 
side  of  AB ;  the  two  angles  will  be  equiva- 
c  lent  to  two  right  angles. 

At  C,  where  the  line  DC  meets  AB,  draw  a  perpendicular  to 
AB  from  C  (I.  '7),  then  the  angles  ACE  and  ECB  are  two  right 
angles  (I.  def.  10).  But  (I.  ax.  10)  the  angle  ECB  is  equivalent 
to  the  angles  ECD  and  DCB  both  together ;  likewise  the  angles 
ACE  and  ECB  together  are  equivalent  to  the  angles  ACD  and 
DCB  both  together;  hence  (I.  ax.  1)  the  angles  ACD  and  DCB 
together  ai-e  equivalent  to  two  right  angles.  Wherefore,  when 
one  straight  line  meets,  etc. 

Cor.  Hence,  if  the  straight  line  DC  be  pi-oduced  on  the  other 
side  of  AB,  the  four  angles  made  by  DC  produced  and  AB 
are  together  equivalent  to  four  right  angles. 

Hence,  also,  all  the  angles  formed  by  any  number  of  straight 
lines  intersecting  one  another  in  a  common  point  are  together 
equivalent  to  four  right  angles. 

Prop.  X. — Theor. — 7)^,  at  a  point  in  a  straight  line,  tico 
other  straight  lines  on  the  ojyposite  sides  make  the  adjacent 
angles  together  equivalent  to  two  right  atigles,  those  two  straight 
lines  are  in  one  and  the  same  straight  line. 

Let  DC  be  the  straight  line  which  makes,  at  the  point  C, 
with  AC  and  CB,  two  adjacent  angles  ACD  and  DCB  togethei- 
equivalent  to  two  right  angles ;  AC  and  CB  are  in  one  and  the- 
same  straight  line. 


—  B 


BOOK  I.]  EUCLID  AND  LEGENDEE.  23 

From  C  draw  a  perpendicular  to  AC  (I.  ^ 

7),  then  the  angle  ACE  is  a  right  angle 
(I.  def.  10).  But  (hyp.)  ACD  and  DCB 
are  together  equivalent  to  two  right 
angles,  so  ACE  and  ECB  are  equivalent  * 
to  two  right  angles  (I.  ax,  1) ;  hence  ACE  being  a  right  angle 
(const,  and  I.  def.  10),  ECB  must  also  be  a  right  angle;  then 
EC  is  perpendicular  to  CB  (I.  def.  10),  and  the  angles  ACE  and 
ECB  are  equal  (I.  ax.  11) ;  therefore  (I.  def  10)  EC  is  a  straight 
line  which  makes  two  equal  angles  with  AB.  But  AC  and  CB 
make  with  EC  the  same  equal  angles ;  hence  AC  and  CB  are 
the  same  straight  line  with  AB.  And  DC  makes  with  AC  and 
CB  (hyp.)  two  adjacent  angles  equivalent  to  two  right  angles, 
but  DC  makes  with  AB  (I.  9)  the  same  angles  equivalent  to 
two  right  angles  ;  hence  AC  and  CB  are  the  same  straight  line 
with  AB  (I.  def  7). 


OTIIER^VISE, 


It  is  proven  (I.  9)  that  the  angles  ACD  and  DCB  are  to- 
gether equivalent  to  two  right  angles  ;  then,  as  C  is  a  point  in 
AB,  AC  can  be  one  line  and  CB  another  (I.  def  7,  scho.) ; 
hence  AC  and  CB  are  in  one  and  the  same  straight  line,  be- 
cause  the  two  unequal  angles  ACD  and  DCB  are  equivalent  to 
two  right  angles  (I.  9).     Wherefore  if,  at  a  point,  etc. 

Prop.  XI. — Theoe. — If  two  straight  lines  cut  one  another^ 
the  vertical  or  opposite  angles  are  equal. 

Let  the  two  straight  lines  AB,  CD  cut  one  another  in  E  ;  the 
angle  AEC  is  equal  to  the  angle  DEB,  and  CEB  to  AED. 

Because  the  straight  line  AE  makes  with  CD  the  angles 
CEA,  AED,  these  angles  are  together  equivalent  (I.  9)  to  two 
risrht  angles.  Again :  because  DE  n 
makes  with  AB  the  angles  AED,  DEB 
these  also  are  together  equivalent  to 
two  right  angles;  and  CEA,  AED  ^ 
have  been  demonstrated  to  be  equivalent  to  two  right  angles ; 
wherefore  (I.  ax.  11  and  1)  the  angles  CEA,  AED  are  equal  to 
the  angles  AED,  DEB.  Take  away  the  common  angle  AED, 
and   (I.  ax.  3)    the  remaining   angles  CEA,  DEB  are  equal 


24  THE   ELEMENTS   OF  [bOOK   I. 

lu  the  same  manner  it  can  be  demonstrated  that  the  angles 
CEB,  AED  are  equal.     Therefore,  if  two  straight  lines,  etc. 

Cor.  If  at  a  point  in  a  straight  line  two  other  straight  lines 
meet  on  the  opposite  sides  of  it,  and  make  equal  angles  with  the 
parts  of  it  on  opposite  sides  of  the  point,  the  two  straight  lines 
are  in  one  and  the  same  straight  line. 

Let  AEB  be  a  straight  line,  and  let  the  angles  AEC,  BED  be 
equal,  CE,  ED  are  in  the  same  straight  line.  For,  by  adding 
the  angle  CEB  to  the  equal  angles  AEC,  BED,  we  have  BED, 
BEC  together  equal  to  AEC,  CEB,  that  is  (I.  9),  to  two  right 
angles ;  and  therefore,  by  this  proposition,  CE,  ED  are  in  the 
same  straight  line. 

Scho.  In  the  proof  here  given,  the  common  angle  is  AED ; 
and  CEB  might  with  equal  propriety  be  made  the  common 
angle.  In  like  manner,  in  proving  the  equality  of  CEB  and 
AED,  either  AEC  or  BED  may  be  made  the  common  angle. 
It  is  also  evident,  that  when  AEC  and  BED  have  been  proved 
to  be  equal,  the  equality  of  AED  and  BEC  might  be  inferred 
from  the  ninth  proposition,  and  the  third  axiom. 

Prop.  XIL — Pkob. — To  describe  a  triangle  of  which  the  sides 
shall  be  equal  to  three  given  straight  lines;  but  any  txoo  of 
these  must  be  greater  than  the  third. 

Let  A,  B,  C  be  three  given  straight  lines,  of  which  any 
two  are  greater  than  the  third ;  it  is  required  to  make  a  tri- 
angle of  which  the  sides  shall  be  equal  to  A,  B,  C,  each  to  each. 
Take  an  unlimited  straight  line  DE,  and  let  F  be  a  point  in 
it,  and  make  FG  equal  to  A,  FH  to  B,  and  HK  to  C.     From 

the  center  F,  at  the  distance  FG, 
describe    (I.    post.    3)    the   circle 
GLM,  and  from  the  center  H,  at 
-E  the   distance    HK,   describe    the 
circle  KLM.   Now,  because  (hyp.) 
g  FK  is  greater  than  FG,  the  cir- 
"  ^  cumference    of   the   circle    GLM 

will  cut  FE  between  F  and  K,  and  therefore  the  circle  KLM 
can  not  lie  wholly  within  the  circle  GLM.     In  like  manner,  be-  , 
cause  (hyp.)  GH  is  gi-onter  than  HK,  the  circle  GLM  can  not 
lie  wholly  witluu  the  ciicle  KLM.     Neither  can  the  circles  be 


BOOK   I.]  ETTCLID   AND    LEGENDRE.  25 

wholly  without  each  other,  since  (hyp.)  GF  and  HK  are  to- 
o-ether  jrreater  than  FH,  The  circles  must  therefore  intersect 
each  other ;  let  them  intersect  in  the  point  L,  and  join  LF,  LH ; 
the  triangle  LFII  has  its  sides  equal  respectively  to  the  three 
lines  A,  B,  C. 

Because  F  is  the  center  of  the  circle  GLM,  FL  is  equal  (I. 
def.  16)  to  FG;  but  (const.)  FG  is  equal  to  A;  therefore  (I.  ax. 
1)  FL  is  equal  to  A,  In  like  manner  it  may  be  shown  that 
HL  is  equal  to  C,  and  (const.)  FH  is  equal  to  B ;  therefore  the 
three  straight  lines  LF,  FH,  HL  are  respectively  equal  to  the 
three  lines  A,  B,  C  ;  and  therefore  the  triangle  LFH  has  been 
constructed,  having  its  three  sides  equal  to  the  three  given  lines, 
A,  B,  C ;  which  was  to  be  done. 

Scho.  It  is  evident  that  if  MF,  MH  were  joined,  another  tri- 
angle would  be  formed,  having  its  sides  equal  to  A,  B,  C,  It 
is  also  obvious  that  in  the  practical  construction  of  this  problem, 
it  is  only  necessary  to  take  with  the  compasses  FH  equal  to  B, 
and  then,  the  compasses  being  opened  successively  to  the 
lenorths  of  A  and  C,  to  describe  circles  or  arcs  from  F  and  H  as 
centers,  intersecting  in  L ;  and  lastly  to  join  LF,  LH. 

The  construction  in  the  proposition  is  made  somewhat  differ- 
ent from  that  given  in  Simson's  Euclid,  with  a  vicAv  to  obviate 
objections  arising  from  the  application  of  this  proposition  in  the 
one  that  follows  it. 

Prop.  XIII. — Prob. — At  a  given  point  in  a  given  straight 
line,  to  make  a  rectilineal  angle  equal  to  a  given  one. 

Let  AB  be  the  given  straight  line,  A  the  given  point  in  it, 
and  C  the  given  angle ;  it  is  required  to  make  an  angle  at  A, 
in  the  straight  line  AB,  that  shall  be  equal  to  C. 

In  the  lines  containing  the  angle  C, 
take  any  points  D,  E,  and  join  them, 
and  make  (L  12)  the  triangle  AFG, 
the  sides  of  which,  AF,  AG,  FG,  shall 
be  equal  to  the  three  straight  lines 

CD,  CE,  DE,  each  to  each.     Then, 
because  FA,  AG  are  equal   to   DC, 

CE,  each  to  each,  and  the  base  FG  to 
the  base  DE,  the  angle  A  is  equal  (I. 


26 


THE   ELEMENTS    OF 


[book 


4)  to  the  angle  C.  Therefore,  at  the  given  jsoint  A  in  the  given 
straight  line  AB,  the  angle  A  is  made  equal  to  the  given  angle 
C ;  which  was  to  be  done. 

Scho.  The  construction  is  easy,  by  making  the  triangles  isos- 
celes. In  doing  this,  arcs  are  described  with  equal  radii  from  C 
and  A  as  centers,  and  their  chords  are  mad«  equal. 

It  is  evident  that  another  angle  might  be  made  at  A,  on  the 
other  side  of  AB,  equal  to  C. 

Prop.  XIV. — Theor. — If  two  angles  of  one  triangle  he  equal 
to  two  angles  of  another^  each  to  each,  and  if  a  side  of  the  one 
he  equal  to  a  side  of  the  other  siniiliarly  situated  with  respect 
to  those  angles  ;  (l)  the  remaining  sides  are  equal,  each  to  each  ; 
(2)  the  remaining  angles  are  equal ;  and  (3)  the  triangles  are 

equcd. 

This  proposition  is  the  converse  of  the  third  proposition,  and 
is  susceptible  of  three  cases,  viz. :  first,  when  the  equal  sides  are 
between  the  equal  angles  ;  secondly  and  thirdly,  when  the  equal 
sides  are  opposite  to  the  equal  angles  similarly  situated. 

Q     J,  Let  ABC  and  DEF  be  two  triangles 

M-hich  have  the  angles  ACB  and  EFD 
equal;  the  angles  BAG  and  DEF 
equal,  and  the  sides  CA  and  FE  equal ; 
then  the  sides  CB  and  FD  are  equal, 
also  the  sides  AB  and  DE ;  the  angles 
CBA  and  FDE  are  equal,  and  the  tri- 
angles are  equal  to  one  another. 

If  the  triangles  be  placed  so  as  to 
have  their  sides  CB  and  FD  in  the  same  straight  line,  but  the 
triangles  be  on  opposite  sides  of  that  line,  and  the  vertex  C  on 

the  vertex  F ;  then  because  the  angles 
AFD  and  EFD  are  equal  (hyp.),  the  angle 
AFE  is  bisected  by  FD ;  and  because  AF 
and  FE  are  equal  (hyp.),  the  triangle  AFE 
is  isosceles  (I.  def  13),  and  the  line  AE 
(T.  6)  is  also  bisected  by  FD;  hence  (I.  1, 
cor.  1)  the  angles  FAE  and  FEA  are 
equal ;  therefore  (I.  3)  the  triangles  FAH 
and  FEII  are  equal.     But  the  angles  FAD 


BOOK   I.]  EUCLID   AND   LEGENDKE.  27 

and  FED  are  eq^^al  (byp.) ;  tlien  taking  from  each  the  equal 
angles  FAII  and  FEH,  there  will  remain  (I.  ax,  3)  the  angle 
HAD  equal  to  the  angle  HED;  hence  (I.  1,  cor.  2)  the  triangle 
AED  is  isosceles,  and  (I.  def.  13  and  I.  cor.  2)  the  side  AD  is 
equal  to  the  side  ED,  and  AE  being  bisected  by  FD  (I.  6),  we 
have  the  triangles  AHD  and  EHD  (I.  3)  equal ;  therefore  the 
angles  HDA  and  HDE  are  equal;  hence  the  triangles  AFD 
and  EFD  have  the  sides  AF  and  EF  (hyp.)  equal,  the  angles 
FAD  and  FED  equal  (hyp.),  and  the  sides  AD  and  DE  equal 
(I.  def  13  and  I,  cor.  2) ;  therefore  (I.  3)  the  angles  FDA  and 
FDE  are  equal ;  the  side  FD  is  common,  and  the  two  triangles 
are  equal. 

In  a  similar  manner,  the  second  and  third  cases  can  be  dem- 
onstrated.    Wherefore,  if  two  angles  of  one  triangle  be,  etc. 

Cor.  Hence,  from  this  proposition,  it  can  also  be  shown  that 
the  second  corollary  to  the  first  proposition  is  true.  Let  the 
triangle  ABC  have  the  angle  CAB  equal  c 

to  the  angle  CBA,  then  will  AC  be  equal 
toCB. 

From  the  vertex  C  (I.  S)  draw  CD  per- 
pendicular to  AB,  then  the  angles  CDA 
and  CDB  are  both  right  angles  (I.  def  10),     ^ 
the  angles  CAD  and  CBD  are  equal  (hyp.),  and  the  side  CD  coni- 
mon  (const.);  therefore  (I.  14)  the  sides  AC  and  CB  are  equal. 

8cho.  It  will  be  seen  from  propositions  third,  fourth,  and 
fourteenth  of  this  book,  that  two  triangles  are  in  every  respect 
equal  when  the  three  sides  of  the  one  are  respectively  equal  to 
the  three  sides  of  the  other,  when  two  triangles  have  two  angles 
and  a  side  in  each  equal,  each  to  each,  and  when  one  angle  and 
two  sides  of  one  are  equal  to  one  angle  and  two  sides  of  the 
other,  each  to  each,  and  the  equal  angles  in  each  triangle  simi- 
larly situated  with  respect  to  those  sides,  but  with  this  limita- 
tion, that  when  two  equal  sides  respectively  of  the  triangles  are 
the  least  sides,  that  the  angles  opposite  the  greater  sides  respec- 
tively of  the  triangles  must  be  of  the  same  kind,  either  both 
acute,  both  right-angled,  or  both  obtuse.  And  from  these  prop- 
ositions it  is  shown  that  of  the  sides  and  angles  of  a  triangle, 
three  must  be  given  to  determine  the  triangle,  and  these  three 
can  not  all  be  angles.     Were  only  three  angles  given,  the  sides, 


28  THE    ELEMENTS    OF  [bOOK   I. 

as  will  appear  from  the  twentieth  prosposltion  of  this  book, 
might  be  of  any  magnitude  whatever.  The  pupil  may  occupy 
himself  in  proving  these  propositions  by  superposition  or 
some  other  way,  as  by  pursuing  a  course  of  demonstration 
different  from  what  is  given  in  the  text,  he  will  more  readily 
familiarize  himself  with  the  process  of  geometrical  reasoning. 

Prop.  XV. — Theor. — Parallel  straight  liries  are  equally/  dis- 
tant from  each  other,  however  so  far  they  be  produced  on  the 
same  plane. 

Let  the  straischt  line  AB 
be  bisected  (I.  6)  at  C,  and 
let  the  perpendicular  CD  be 
drawn  (I.  7) ;  join  AD  and 
A  c  '     ^      BD,  and  if  the  triangle  ADC 

be  applied  to  the  triangle  BDC  so  that  they  will  fall  on  difler- 
€nt  sides  of  BD  and  have  D  and  BD  common,  their  sides  DE 
and  CB  will  be  equally  distant  from  each  other,  and  so  they 
will  never  meet,  however  so  far  they  be  produced  on  the  same 
plane,  and  consequently  (I.  def  11)  are  parallel  straight  lines. 

Because  DC  is  perpendicular  to  AB  (const.),  the  angles  ACD 
and  BCD  are  both  right  angles  (I,  def  10),  and  are  equal  (I.  ax. 
11)  ;  hence  the  triangles  ADC  and  BDC  have  the  side  DC 
common,  the  sides  AC  and  CB  equal  (const.),  and  the  angles 
ACD  and  BCD  equal  (I.  def  10  and  ax.  11);  therefore  (I.  3) 
the  triangles  are  equal,  having  the  sides  AD  and  BD  equal ; 
the  angle  ADC  equal  to  the  angle  BDC,  and  the  angle  DAC 
equal  to  the  angle  DBC.  Now,  when  ADC  is  applied  to  BDC 
so  that  they  will  fall  on  difterent  sides  of  BD,  and  have  D  and 
BD  common  (hyp.),  since  AD  is  equal  to  BD,  the  point  A  will 
fall  on  B ;  hence  the  angle  EBD  will  be  the  same  as  ADC,  and 
equal  to  BDC,  the  angle  BED  the  same  as  ACD,  and  equal  to 
DCB,  the  angle  EDB  the  same  as  DAC,  and  equal  to  DBC,  the 
side  DE  the  same  as  AC,  and  equal  to  AC ;  the  side  EB  equal 
to  DC;  therefore  the  triangles  BDC  and  BDE  are  equal  (I. 
ax.  8),  having  their  sides  and  angles  equal,  each  to  each. 

Since  DE  and  CB  are  straight  lines,  they  have  no  variation 
in  the  direction  of  their  lengths  from  D  to  E  or  from  C  to  B 
(I.  def  7) ;  and  because  DC  is  equal  to  EB,  DE  and  CB  are 


BOOK  I.]  EUCLID  AND  LEGENDEE.  29 

equally  distant  from  each  other  at  their  extremities,  and  having 
no  variation  in  the  direction  of  their  lengths  from  D  to  E  and 
from  C  to  B  (I.  def.  7),  they  are  also  equally  distant  from  each 
other  at  every  part  between  D  and  E  and  C  and  B,  each  to 
each ;  therefore  DE  and  CB  (I.  post.  2)  on  being  produced  to 
any  length  are  still  the  same  straight  lines,  and  will  have  no 
variation  in  the  direction  of  their  lengths  (I.  def.  7),  conse- 
quently they  will  always  be  the  distant  DC  or  EB  from  each 
other,  at  every  part,  each  to  each ;  and  being  always  the  same 
distant  DC  or  EB  from  each  other,  will  never  meet,  and  are 
parallel  straight  lines  (I.  def  11).  Wherefore,  parallel  straight 
lines,  etc. 

Cor.  1.  In  like  manner,  it  can  be  shown  that  DC  and  EB 
are  parallel  lines;  hence  DCBE  is  a  parallelogram  (I.  def  14) ; 
and  since  the  angles  DCB  and  DEB  are  equal,  and  the  angles 
EDB  and  BDC  equal  to  the  angles  EBD  and  DBC  (I.  ax.  2), 
and  the  sides  DE  and  CB  equal,  also  the  sides  CD  and  BE 
equal,  a  parallelogram  has  its  opposite  sides  and  opposite  angles 
equal. 

Cor.  2.  Hence  parallel  lines,  DE  and  CB,  intercepted  by  a 
straight  line,  DB,  make  the  alternate  angles  EDB  and  CBD 
equal;  and,  co?it»er5eZy,  when  the  alternate  angles  EDB  and  CBD 
are  equal,  the  lines  DE  and  CB  are  parallel. 

Cor.  3.  In  the  parallelogram,  the  opposite  sides  being  equal, 
the  straight  lines  which  join  the  extremities  of  two  equal  and 
parallel  straight  lines  toward  the  same  parts — that  is,  the  near- 
est extremities  together — are  themselves  equal  and  parallel ; 
hence  a  quadrilateral  which  has  two  sides  equal  and  parallel  is 
(I.  def.  14)  a  parallelogram. 

Cor.  4.  Because  the  triangles  DCB  and  DEB  are  equal,  a 
diagonal,  DB,  bisects  the  parallelogram ;  and  if  two  parallelo- 
grams have  an  angle  of  the  one  equal  to  an  angle  of  the  other, 
and  the  sides  containing  those  equal  angles  respectively  equal, 
the  parallelograms  are  equal,  as  the  parallelograms  can  be  bi- 
sected by  diagonals  subtended  by  the  equal  angles,  and  the  tri- 
angles thus  formed  are  equal  (I.  3) ;  hence  (I.  ax.  6)  the  par- 
allelograms are  equal;  hence,  also,  if  a  pai'allelograra  and  a  tri- 
angle be  upon  the  same  or  equal  bases,  and  between  the  same 
parallels,  the  parallelogram  is  double  the  triangle. 


30  THE   ELEMENTS   OF  [bOOK  I. 

Cor.  5.  Hence,  also,  parallelograms  upon  the  same  or  equal 
bases,  and  between  the  same  parallels,  are  equal ;  and  triangles 
upon  the  same  or  equal  bases  and  between  the  same  parallels, 
are  equal. 

Cor.  6.  Hence,  from  the  preceding  corollary,  it  is  plain  that, 
triangles  or  parallelograms  between  the  same  parallels,  but 
upon  unequal  bases,  are  unequal. 

Cor.  T.  And  a  straight  line  drawn  from  the  vertex  of  a  tri- 
angle  to  the  point  of  bisection  of  the  base,  bisects  the  triangle ; 
and  if  two  triangles  have  two  sides  of  the  one  respectively  equal 
to  two  sides  of  the  other,  and  the  contained  angles  supple- 
mental (I.  def  20),  the  triangles  are  equivalent;  the  converse  is 
also  true. 

Cor.  8.  If  through  any  point  in  either  diagonal  of  a  parallelo- 
gram straight  lines  be  drawn  parallel  to  the  sides  of  the  four 
parallelograms  thus  formed,  those  through  which  the  diagonal 
does  not  pass,  and  which  are  called  the  com2)lements  of  the 
other  two,  are  equivalent. 

Peop.  XVI. — Theoe. — If  a  straight  linefallupon  tioo  parallel 
straight  Ihies,  (l)  it  makes  the  alternate  angles  equal  to  one  an- 
other ;  (2)  the  exterior  angle  equal  to  the  interior  and  remote 
upon  the  same  side,  and  (3)  the  tico  interior  angles  upon  the 
same  side  together,  equivalent  to  two  right  angles. 

Let  the  straight  lines  AB,  CD  be  parallel,  and  let  EF  fall  upon 
them;  then  (l)  the  alternate  angles  AGH,  GHD  are  equal  to 
one  another ;  (2)  the  exterior  angle  EGB  is  equal  to  the  interior 
and  remote  upon  the  same  side,  GHD;  and  (3)  the  two  in- 
terior angles  BGH,  GHD  ujDon  the  same  side  are  together 
equivalent  to  two  right  angles. 

Since  AB,  CD  are  parallel  (hyp.),  theper- 

B  pendiculars  (I.  7)  MG,  LH  make  the  angles 

MGH,  GIIL  equal  to  one  another  (I.  15, 

cor.   2),  and  AG3I,  LHD   are  two  riglit 

.jj  angles  (I.   def   10);  if  Ave  add  AGM  to 

MGH  they  will  be  equal  to  AGH  (I.  ax. 

^'      10),  and  DHL  added  to  LIIG  are  likewise 

equal  to  GIID ;  hence  (I.  ax.  2)  AGH  and  GHD  are  equal  to 

one  another. 


BOOK  I.]  EUCLID  AND  LEGENDRE.  31 

Second.  AGII  is  equal  to  EGB  (I.  11),  therefore  (I.  ax,  1) 
EGB  is  equal  to  GHD. 

Third.  Add  to  EGB  and  GHD,  each,  the  angle  BGII ;  there- 
fore (I.  ax.  2  and  ax.  10)  EGB  and  BGH  are  equivalent  to  the 
angles  GHD  and  BGH,  but  EGB  and  BGH  are  equivalent  to 
two  right  angles  (I.  9) ;  therefore,  also,  BGH  and  GHD  are  to- 
gether equivalent  to  two  right  angles.  Wherefore,  if  a  straight 
line,  etc. 

Cor.  1.  Hence,  conversely,  tAvo  straight  lines  are  parallel  to 
one  another,  if  another  straight  line  falling  on  them  (1)  makes 
the  alternate  angles  equal ;  (2)  the  exterior  angle  equal  to  the 
interior  and  remote  upon  the  same  side  of  that  line ;  and  (3) 
the  two  interior  angles  upon  the  same  side  together  equivalent 
to  two  right  angles. 

Let  EF  fall  on  AB  and  CD,  the  perpendiculars  (I.  1)  GM  and 
HL  make  two  riglit  angles  (I.  def.  10),  AGM  and  DHL.  But 
AGH  and  DHG  are  equal  (hyp.) ;  hence  (L  ax.  3)  MGH  is 
■equal  to  LHG.  But  MGL  and  LHM  are  both  right  angles 
(const,  and  L  def.  10) ;  hence  (I.  ax.  3)  HGL  is  equal  to  GHM; 
•therefore  in  the  triangles  HMG  and  GLH  we  have  two  angles 
in  one  equal  to  two  angles  in  the  other,  each  to  each,  and  the 
side  GH  common,  and  (I.  14)  the  triangles  are  equal;  hence 
GM  and  HL  are  equal,  and  (L  15)  AB  and  CD  are  equally  dis- 
tant from  each  other,  and  will  never  meet  on  being  produced 
{L  post.  2),  and  are  parallel  (L  def'  11). 

Because  EGB  is  equal  to  DHG  (hyp.),  and  EGB  equal  to 
AGH  (I.  11),  the  angle  AGH  is  equal  to  DHG  (L  ax.  1),  but 
they  are  alternate  angles;  therefore  (L  16,  cor.  1,  part  l)  AB  is 
parallel  to  CD.  Again :  because  BGH  and  GHD  are  together 
equivalent  to  two  right  angles  (hyp.),  and  AGH  and  BGH  are 
also  equivalent  to  two  right  angles  (I.  9),  the  angles  AGH  and 
BGH  are  together  equal  to  BGH  and  GHD  (I.  7  and  ax.  1) ; 
then  (L  ax.  3)  AGH  is  equal  to  GHD,  but  they  are  alternate 
angles,  therefore  (I.  16,  cor.  1,  part  l)  AB  and  ED  are  parallel. 
Wherefore,  two  straight  lines  are  parallel,  etc. 

Cor.  2.  When  one  angle  of  a  parallelogram  is  a  right  angle, 
all  the  other  angles  are  right  angles;  for  since  (I.  10)  BGM  and 
GMH  are  together  equivalent  to  two  right  angles,  if  one  of 
them  be  a  right  angle,  the  other  must  also  be  a  right  angle,  and 


32  THE   ELEMENTS   OF  [boOK   I. 

(I.  15,  cor.  1)  the  opposite  angles  are  equal.     A  rectangle,  then 
(L  def.  15),  has  all  its  angles  right  angles. 

Cor.  3.  If  two  straight  lines  make  an  angle,  two  others  par- 
allel to  them  contain  an  equal  or  supplemental  angle ;  thus  LGM, 
and  the  vertical  angle  produced  by  AB  and  the  continuance  of 
MG  through  G,  are  each  equal  to  LHM,  while  the  angle  AGM, 
and  its  vertical  angle  contained  by  GB  and  the  continuance  of 
MG,  are  each  equal  to  the  supplement  of  LHM ;  hence  we  can 
divide  a  given  straight  line  AB  into  any  proposed  number  of 
equal  parts. 

Prop.  XVH. — Theor. — Two  straight  lines  which  are  not 
in  the  same  straight  line,  and  which  nre  parallel  to  a  third 
straight  line,  are  parallel  to  one  another. 

Let  the  straight  lines  AB,  CD  be  each  of  them  parallel  to 
the  straight  line  EF ;  AB  is  also  parallel  to  CD. 

Let  the  straight  line  LH  cut  AB,  CD,  EF ;  and  because  LH 
L     cuts  the  parallel  straight  lines  AB,  EF,  the 
angle  LGB  is  equal  (L  16,  part  2)  to  the 
.  J}  angle  LHF.     Again  :  because  the  straight 
line  LH  cuts  the  parallel  straight  lines  CD, 
"  ^  EF,  the  angle  LKD  is  equal  (L 1 6,  part  2)  to 

^  ■        " —  J.  the  angle  LHF ;  and  it  has  been  shown 

that  the   angle   LGB  is  equal  to  LHF; 

wherefore,  also,  LGB  is  equal  (L  ax.  1)  to 

LKD,  the  interior  and  remote  angle  on  the  same  side  of  LH ; 

therefore  AB  is  parallel  (L  10,  part  1)  to  CD.     Wherefore,  two 

straight  lines,  etc. 

Prop.  XVIIL — Prob. — To  draw  a  straight  line  parallel  to  a 
given  straight  line  through  a  given  poi^it  without  it. 

Let  AB  be  the  given  straight  line,  and  C  the  given  point ;  it 
is  required  to  draw  a  straight  line  through  C,  parallel  to  AB. 

In  AB  take  any  point  D,  and  join  CD ;  at  the  point  C,  in  the 

c  straight  line  CD,  make  (I.  13)  the  angle 

E  — -^  F  p^j,  ^^^^^^  ^^  ^j^j,  ^  ^^^  produce  the 

A  ^ B  Straight  line  EC  to  any  point  F. 

Because  (const.)  the  straight  line  CD, 
which  meets  the  two  straight  lines  AB,  EF,  makes  the  alternate 


BOOK  I.]  EUCLID  AND  LEGENDEE.  33 

angles  ECD,  CDB  equal  to  one  another,  EF  is  parallel  (I. 
15,  eor,  2)  to  AB.  Therefore  the  straight  line  EOF  is  drawn 
through  the  given  point  C  parallel  to  the  given  straight  linev 
AB ;  which  was  to  be  done.  • 

Prop.  XIX, — Theok. — If  a  straight  line  meet  two  otho'^ 
straight  lines  xohich  are  in  the  same  pkoie,  so  as  to  make  the 
two  interior  angles  on  the  same  side  of  it,  taken  together,  less 
than  two  right  angles,  these  straight  lines  shall  at  length  meet 
upon  that  side,  if  they  he  continually  i^i'oduced.  Axiom 
twelfth.  Elements  of  Euclid. 

Let  EF  be  the  straight  line  meeting  AB,     ^ 
CD  on  the  same  plane,  so  tliat  BLO,  DOL 
are  less  than  two  right  angles,  the  lines 
AB,  CD  will  meet  if  continually  produced. 

At  the  point  O,  draw  GH  (I.  1 8)  parallel  g 
to  AB,  then  BLO,  HOL  are  equivalent  to 
two  right  angles  (L  IG),  but  DOL  is  less  c 

than  HOL  (I.  ax.  9) ;  therefore  BLO,  DOL  are  less  than  BLO, 
HOL,  and  less  than  two  right  angles.  But  GH,  which  fornij^ 
with  EF  the  angle  HOL,  is  parallel  (const.)  with  AB,  which! 
forms  with  EF  the  angle  BLO,  and  (L  def  11)  GH  and  AB- 
never  meet  each  other,  because  (L  15)  they  are  equally  distant 
from  each  other,  and  (L  16,  cor.  l)  the  interior  angles  HOL^ 
BLO  together  equivalent  to  two  right  angles ;  therefore.^ 
then,  CD,  which  forms  Avith  EF  the  angle  DOL  less  than  HOL, 
can  not  be  parallel  with  AB  (L  ax.  9) ;  and  not  being  parallel 
wath  AB,  CD  and  AB  can  not  preserve  the  condition  of  par- 
allel lines  (L  def  11),  and  will  meet.  And  the  line  CD  making 
with  EF  the  angle  DOL  less  than  HOL,  on  the  side  of  EF, 
where  the  interior  angles  BLO,  DOL  are  less  than  two  ricrht 
angles,  the  line  OD  must  therefore  be  between  GH  and  AB, 
and  is  consequently  nearer  to  AB  than  GH  is  to  AB  (L  ax.  9) ; 
but  CD  making  with  EF  the  angle  EOC  greater  than  the  angle 
GOE,  which  GH  makes  with  EF,  therefore  CO  must  be  with- 
out AB  and  GH,  and  consequently  is  farther  from  AB  than  GH 
is  from  AB  ;  hence  the  straight  line  CD,  Avhich  is  made  of  CO 
and  OD,  has  parts  unequally  distant  from  AB ;  therefore  since 
CD  approaches  AB  on  the  side  of  EF  where  OD  is,  CD  must 
3 


S^  THE   ELEMENTS    OF  [bOOK   I. 

meet  AB  on  that  side  wliea  continuallj^  produced,  but  OD 
makes  the  angle  DOL  less  than  HOL,  therefore  CD  will  meet 
AB  on  the  side  of  EF  where  the  angles  BLO,  DOL  are  less 
than  BLO,  HOL.     Wherefore,  if  a  straight  line,  etc. 

Cor.  1.  Hence  a  straight  line  which  intercepts  one  of  two  or 
more  parallel  straight  lines  will  intercept  the  others  if  continu- 
ally produced ;  hence,  also,  two  straight  lines  which  intercept 
each  other  are  not  both  parallel  to  the  same  straight  line. 

Prop.  XX. — Theor. — If  a  side  of  any  triangle  be  produced, 
(1)  the  exterior  angle  is  equivalent  to  the  tico  interior  and  re- 
mote angles  ;  and  (2)  the  three  interior  angles  of  every  triangle 
are  together  equivalent  to  two  right  angles. 

Let  ABC  be  a  triangle,  and  let  one  of  its  sides  BC  be  pro- 
educed  to  D;  (1)  the  exterior  angle  ACD  is  equivalent  to  the 
two  interior  and  remote  angles  CAB,  ABC;  and  (2)  the  three 
interior  angles,  ABC,  BCA,  CAB,  are  together  equivalent  to 
two  right  angles. 

Through  tlae  point  C  draw  (L  18)  CE  parallel  to  the  straight 

line  AB.     Then,  because  AB  is  parallel  to  EC,  and  AC  falls 

'>iipon  them,  the  alternate  angles  BAC,  ACE  are  (L  16,  part  1) 

equal  Again:  because  AB  is  ijarallel  to 
EC,  and  BD  falls  upon  them,  the  exterior 
angle  ECD  is  equivalent  (L  I G,  part  2) 
to  the  interior  and  remote  ano-le  ^VBC  : 

'B^ ^ — D   but  the  angle  ACE  has  been  sliown  to 

F  be  equal  to  the  angle  BAC  ;  there- 
fore the  whole  exterior  angle  ACD  is  equivalent  (I.  ax.  2)  to 
the  two  interior  and  remote  angles  CAB,  ABC.  To  these 
equals  add  the  angle  ACB,  and  the  angles  ACD,  ACB  are 
equivalent  (L  ax.  2)  to  the  three  angles  CBA,  BAC,  ACB ;  but 
the  angles  ACD,  ACB  are  equivalent  (I.  9)  to  two  right  angles; 
therefore,  also,  the  angles  CBA,  BAC,  ACB  are  equivalent  to 
two  right  angles.     Wherefore,  if  a  side,  etc. 

Another  proof  of  this  important  proposition  can  be  given  by 
producing  the  side  AC  through  C  to  F.  !N"ow  (I.  11),  the  angle 
DCF  is  equal  to  the  angle  ACB ;  EC  being  parallel  (const.)  to 
BA,  the  exterior  angle  ECD  is  equal  to  the  intei'ior  and  remote 
angle  ABC  (L  10),  and  for  the  same  reason  the  alternate  angles 


r 


BOOK  I.J  EUCLID  AND  LEGENDRE.  35 

EGA  and  CAB  are  equal;  hence  we  have  the  three  angles  of 
the  triangle,  ACB,  CBA,  and  CAB,  equal  to  the  three  angles 
DCF,  DCE,  and  EGA,  each  to  each ;  but  (I.  9)  tlie  angles  DCF, 
DCE,  and  ECA  are  equivalent  to  two  right  angles ;  therefore 
(I.  ax.  1)  the  three  angles  ACB,  CBA,  and  CAB  of  the  triangle 
are  likewise  equivalent  to  two  right  angles.  And  when  a  par- 
allelogram (I.  def  14)  is  formed  by  drawing  (I.  18)  pai-allels  to 
BA  and  AC  respectively,  it  can  be  shown  by  the  sixteenth 
proposition  tliat  two  adjacent  angles  of  the  parallelogram  are 
equivalent  to  two  right  angles,  and  the  four  angles  together 
equivalent  to  four  right  angles;  since  (I.  15,  cor.  4)  a  diagonal 
bisects  the  parallelogram  and  forms  two  equal  triangles,  the 
angles  are  also  equally  divided,  hence  each  triangle  has  its 
three  angles  equivalent  to  two  right  angles. 

Cor.  1.  All  tlie  interior  angles  of  any  rectilineal  figure,  to- 
gether Avith  four  right  angles,  are  equivalent  to  twice  as  many 
right  angles  as  the  figure  has  sides. 

For  any  rectilineal  figure  can  be  divided  into  as  many  tri- 
angles as  the  figure  has  sides,  by  draAving  straight  lines  from  a 
point  within  the  figure  to  each  of  its  angles;  and  by  the  prop- 
osition, all  the  angles  of  these  triangles  are  equivalent  to  twice 
as  many  right  angles  as  there  are  triangles — that  is,  as  there  are 
rsides  of  the  figure ;  and  the  same  angles  are  equivalent  to  the 
angles  of  the  figure,  together  with  the  angles  at  the  point  Avhich  is 
the  common  vertex  of  the  triangles — that  is  (I.  9,  cor.),  together 
with  four  right  angles.  Tlierefore  all  the  angles  of  the  figure, 
together  Avitli  four  i-ight  angles,  are  equivalent  (I.  ax.  1)  to 
twice  as  many  riglit  angles  as  the  figure  has  sides. 

Scho.  1.  Another  proof  of  this  corollary  may  be  obtained  by 
■dividing  the  figure  into  triangles  by  lines  drawn  from  any 
.ano-les  to  all  the  remote  angles.  Then  each  of  the  tAVO  ex- 
treme  triangles  has  tAVO  sides  of  the  polygon  for  tAvo  of  its  sides, 
while  each  of  the  other  triangles  has  only  one  side  of  the  figure 
for  one  of  its  sides  ;  and  hence  the  number  of  triangles  is  less 
by  tAVO  than  the  number  of  the  sides  of  the  figure.  Biit  the  in- 
terior angles  of  the  figure  are  evidently  equivalent  to  all  the 
interior  angles  of  all  the  triangles — that  is,  to  tAvice  as  many 
right  angles  as  there  are  triangles,  or  tAvice  as  many  right 
angles  as  the  figure  has  sides,  less  the  angles  of  two  triangles — 


36  THE   ELEMENTS    OP  [bOOK    I. 

tlfet  is,  four  riglit  angles.  Hence,  iu  any  equiangular  figure, 
the  number  of  the  sides  being  known,  the  magnitude  of  each 
angle  compared  with  a  right  angle  can  be  determined.  Thus, 
in  a  regular  pentagon,  the  amount  of  all  the  angles  being  twice 
five  *right  angles  less  four— that  is,  six  right  angles,  each  angle 
will  be  one  fifth  j^art  of  six  right  angles,  or  one  right  angle  and 
one  fifth.  In  a  similar  manner  it  would  appear  that  in  the 
regular  hexagon,  each  angle  is  a  sixth  part  of  eight  right  angles, 
or  a  right  angle  and  a  third ;  that  in  the  regular  heptagon,  each 
is  a  right  angle  and  three  sevenths  ;  in  the  regular  octagon,  a 
rio-ht  ano;le  and  a  half,  etc. 

Cor.  2.  All  the  exterior  angles  of  any  rectilineal  figure  are 
together  equivalent  to  four  right  angles. 

Because  each  interior  angle  ABC,  and  the  adjacent  exterior 
ABD,  are  together  equivalent  (I.  9)  to 
two  right  angles,  therefore  all  the  in- 
terior, together  with  all  the  exterior 
angles  of  the  figure,  are  equivalent  to 
twice  as  many  right  angles  as  there  are 
B  sides  of  the  figure — that  is,  hj  the  fore- 

going corollary,  they  are  equivalent  to  all  the  interior  angles 
of  the  figure,  together  with  four  right  angles;  therefore  all  the 
exterior  angles  are  equivalent  (I.  ax.  .3)  to  four  right  angles. 

Sc/to.  2.  It  is  to  be  observed,  that  if  angles  be  taken  in  the 
ordinary  meaning,  as  understood  by  Euclid,  this  corollary  and 
the  foregoing  are  not  applicable  when  the  figures  have  re-entrant 

angles — that  is,  such  as  open  outward. 
The  second  corollary  will  hold,  how- 
ever, if  the  difference  between  each 
re-entrant  angle  and  tAvo  riiiht  ansrles 
be  taken  from  the  sum  of  the  other  exterior  anccles:  and  the 
former  will  be  applicable,  if,  instead  of  the  angle  which  opens 
externally,  the  difterence  between  it  and  four  right  angles  be 
used.  Both  corollaries,  indeed,  Avill  hold  without  change,  if  the 
re-entrant  angle  be  regarded  as  internal  and  greater  than  two 
right  angles;  and  if,  to  find  the  exterior  angles,  the  interior  be 
taken,  in  the  algeljraic  sense,  from  two  right  angles,  as  in  this 
case,  the  re-entering  angles  will  give  negative  or  subtractive 
results. 


BOOK  I.]  EUCLID  AND  LEGENDEE.  87 

Cor.  3.  If  a  triangle  has  a  right  angle,  the  remaining  angles 
are  together  eqiiivaleut  to  a  right  angle ;  and  if  one  angle  of  a 
triangle  be  equivalent  to  the  other  two,  it  is  a  right  angle. 

Cor.  4.  The  angles  at  the  base  of  a  right-angled  isosctles  tri- 
angle are  each  half  a  right  angle. 

Cor.  5.  If  two  angles  of  one  triangle  be  equal  to  two  angles 
of  another,  their  remaining  angles  are  equal. 

Cor.  6.  Each  angle  of  an  equilateral  triangle  is  one  third  of 
two  right  angles,  or  two  thirds  of  one  right  angle. 

Cor.  1.  Hence,  a  right  angle  may  be  trisected  by  describing 
an  equilateral  triangle  on  one  of  the  lines  containing  the  right 
angle. 

ScJio.  3.  By  this  principle  also,  in  connection  with  the  fifth 
proposition,  we  may  trisect  any  angle,  Avhich  is  obtained  by  the 
successive  bisection  of  a  right  angle,  such  as  the  half,  the 
fourth,  the  eighth,  of  a  right  angle,  and  so  on. 

Cor.  8.  Any  two  angles  of  a  triangle  are  less  than  two  right 
angles. 

Cor.  9.  Hence  every  triangle  must  have  at  least  two  acute 
angles. 

Peop.  XXI.  — Tiieor. — Iftxco  sides  of  a  triangle  be  unequal, 
(1)  the  greater  side  has  the  greater  angle  opposite  to  it  ;  and  (2) 
conversely,  if  tioo  angles  of  a  triangle  be  unequal,  the  greater 
angle  has  the  greater  side  opjiosite  to  it. 

D  Let  ABC  be  a  triangle  of  which  the  side 

AB  is  greater  than  the  side  AC ;  the  angle 
ACB  opposite  AB  is  greater  than  the  angle 
ABC  opposite  AC. 

Because  AB  is  greater  than  AC,  pro- 
duce AC  (I.  post.  2),  and  with  A  as  a  center,  and  a  radius  AB, 
describe  a  circle  (I.  post.  3)  intercepting  AC  produced  in  D, 
and  join  BD ;  the  triangle  ADB  is  isosceles  (I.  defs.  13  and  16) ; 
therefore  the  angle  ADB  is  equal  to  the  angle  ABD  (I.  1,  cor.  1). 
But  (I.  20)  the  exterior  angle  ACB  is  equivalent  to  the  sum  of 
the  two  remote  interior  angles  CDB  and  DBC.  And  CDB  is 
equal  to  DBA  (I.  1,  cor.  1),  but  ABC  is  less  than  ABD  (I.  ax. 
9) ;  therefore  ACB  is  greater  than  CBA. 

The  proof  can  also  be  given  by  laying  oflf  on  the  greater  side 


38  THE    ELEMENTS    OF  [bOOK   I. 

AB,  tlie  less  side  AC ;  joining  the  vertex  of  the  opposite  angle 
with  the  point  where  AC  terminates  on  AB ;  and  the  demon- 
stration conducted  similarly  to  the  preceding.  Again :  by 
cutting  oflf  on  AC  a  part  equal  to  CB,  hisect  the  angle  BCA, 
and  join  the  extremity  of  the  part  on  AC  equal  to  CB  with  the 
foot  of  the  line  bisecting  the  angle  BCA ;  this  proof  is  given  by 
means  of  (I.  3  and  20). 

Conversely:  when  the  angle  ACB  is  greater  than  the  angle 
ABC,  the  side  AB  is  greater  than  AC.     At  the  vertex  of  BCA 

on  AC  make  an  angle  ACD  equal 
to  the  angle  ABC  (I.  13).  Now, 
the  angle  ACD,  the  equal  to  the 
angle  ABC,  is  subtended  by  AD, 
~^  ^  ^  and  the  angle  ACB  is  subtended  by 

AB,  but  AB  is  greater  than  AD ;  hence  the  greater  angle  is 
subtended  by  the  greater  line  ;  therefore,  in  the  triangle  ACB, 
the  greater  angle  ACB  is  subtended  by  a  greater  side  than  the 
less  angle  ABC.  And  (L  def.  19)  if  the  angles  ACD  and  ABC 
be  subtended  by  arcs,  the  arc  subtending  ACB  is  greater  than 
the  arc  subtending  ABC ;  but  the  side  AB  is  the  chord  of  the 
arc  subtendins:  ACB,  and  AC  is  the  chord  of  the  arc  subtending 
ABC ;  therefore  AB  is  greater  than  AC.  Wherefore,  if  two 
sides  of  a  triangle,  etc. 

Cor.  Hence  any  two  sides  of  a  triangle  are  together  greater 
than  the  remaining  side. 

Scho.  The  truth  of  this  corollary  is  so  manifest,  that  it  is 
given  as  a  corollary  to  avoid  increasing  the  number  of  axioms. 
Archimedes  defined  the  straight  line  the  shortest  distance  be- 
tween two  points ;  hence  two  straight  lines  connecting  three 
points  not  in  the  same  direction,  are  together  greater  than  one 
straight  line  connecting  any  two  of  those  jDoiuts. 

Prop.  XXII, — Tiieor. — If  two  triangles  have  tico  sides  of 
the  one  equal  to  two  sides  of  the  other,  each  to  each,  hut  the 
angles  contained  hg  those  sides  loiequal,  the  base  or  remaining 
side  of  the  one  ichich  has  the  greater  angle  is  greater  than  the 
base  or  reraaining  side  of  the  other. 

Let  DEG  and  DEF  be  two  triangles  which  have  the  sides 
DE  common,  the  sides  DG  and  DF  equal,  but  the  angle  EDG 


HOOK  I.]  EUCLID  AND  LEGENDEE.  39 

greater  tlian  the  angle  EDF  ;  the  side  EG  is  also  greater  than 
the  side  EF. 

In  the  triang-les  DEF  and  DEG  Ave  have 
(hyp.)  DE  commoii,  DG  equal  to  DF,  and 
the   an'He   EDG   crreater    than    the    anHe 
EDF.     Now,  because  DG  and  DF  are  equal, 
the  angles  DGF  and  DFG  are  equal  (I.  1, 
cor.  1) ;  but  the  angle  DGF  is  greater  than 
the  angle  EGF  (I.  ax.  9) ;  therefore  the  angle 
DFG  is  greater  than  the  angle  EGF ;  and  much  more  is  the 
angle  EFG  greater  than  the  angle  EGF.     Then  (I.  21),  EG  op- 
posite EFG  is  greater  than  EF  opposite  EGF.     Wherefore,  if 
two  triangles  have,  etc. 

There  are  other  cases  of  this  proposition ;  if  a  line  equal  to 
DE,  the  less  side,  be  drawn  through  D,  making  with  DF,  on  the 
same  side  of  it  with  DE,  an  angle  equal  to  EDG,  the  extremity  of 
that  line  might  fall  on  FE  produced,  or  above,  or  helo^o  it.  Or 
the  ano-le  DFE  could  be  in  the  triangle  DEG,  or  on  the  other 
side  of  DE.  And  a  very  easy  proof  can  be  given  by  bisecting 
the  angle  FDG  by  a  straight  line  cutting  EG  in  a  point,  which 
call  K,  and  joining  DK  and  KF,  for  KG  would  be  equal  to 
KF  (I.  3) ;  adding  EK,  we  would  have  EG  equivalent  to  EK 
and  KF;  therefore  (I.  21,  cor.)  EG  greater  than  EF. 

Cor.  Hence,  conversely,  if  two  triangles  have  two  sides  of 
the  one  equal  to  two  sides  of  the  other,  each  to  each,  but  their 
bases  unequal,  the  angles  contained  by  the  respectively  equal 
sides  of  those  triangles  arc  also  unequal,  the  greater  angle 
being  in  the  triangle  which  has  the  greater  base.  For  with  a 
radius,  DG,  or  its  ecpial,  DF,  describe  a  circle  (I.  post.  3)  from  a 
center,  D,  and  draw  on  the  same  side  of  DE  with  the  angle 
EDF  a  line  equal  to  EG  from  the  other  extremity  of  DE  to  the 
circumference,  then  in  the  triangles  DEG  and  DEF  we  have 
(const.)  DG  equal  to  DF  (I.  def  16).  But  (hyp.)  the  side  EF 
is  less  than  EG ;  hence  the  angle  EDF  is  less  than  the  angle 
EDG  (I.  ax.  9). 

Pkop.  XXin. — Peob. — To  describe  a  parallelogram  upon  a 
given  straight  line. 

Let  DB  be  a  given  straight  line ;  it  is  required  to  describe  a 


40 


THE    ELEMENTS    OF 


[book  I. 


paralleloGjram.  U])Ou  it.     From  a  point,  D,  on  DB  draw  DF  (I. 


F 


H 


B 


post.  1),  then  from  the  point  F  on 
DF  draw  FL  parallel  to  DB  (I.  18) ; 
and  if  through  B,  a  point  on  DB,  a 
parallel  to  DF  (I.  18),  he  drawn, 
DBFG  (I.  def.  15)  is  a  parallelogram. 
If  from  D  a  perpendicular,  DII  (I.  7),  be  drawn,  then  the 
angle  HDB  is  a  riglit  angle  (I.  def  10)  ;  and  from  H  a  parallel 
to  DB  as  HL  be  drawn  (I.  18),  and  from  B  a  parallel  to  HD 
as  LB  be  drawn  (I.  18),  then  DBIIL  (I.  def  15)  is  a  rectangle. 
And  if  from  D  as  a  center,  and  a  radius  DB  (I.  post.  3),  an  arc 
be  described  intercepting  DH,  or  DH  produced,  and  a  rectangle 
be  described  from  that  point  where  DH  is  intercepted,  that 
rectangle  will  be  a  square  (I.  def  15). 

Co?:  1.  Hence  (const,  and  I.  15,  cor.  l)  a  square  has  all  its 
sides  equal,  and  (I.  15,  cor.  l)  all  its  angles  are  right  angles. 

Cor.  2.  Hence  the  squares  described  on  equal  straight  lines 
(I.  15,  cor.  4)  are  equal. 

Cor.  3.  If  two  squares  be  equal,  their  sides  are  equal   (I. 
ax.  1). 

Cor.  4.  If  AB  and  AD,  two  adjacent  sides  of  a  rectangle 
BD,  be  divided  into  parts  which  are  all  equal,  straight  lines 
drawn  through  the  points  of  section,  parallel  to  the  sides,  divide 
the  rectangle  into  squares  which  are  all  equal,  and  the  number 
A  B  of  which  is  equal  to  the  product  of  the 

number  of  parts  in  AB,  one  of  the  sides, 
multiplied  by  the  number  of  parts  in 
AD,   the    other.       For    (const.)    these 


figures  are  all  parallelograms;  and  (I. 
c  def.  15  and  const.)  the  sides  being  equal, 
and  the  angles  being  (I.  16,  part  2)  equal  to  A,  and  there- 
fore right  angles,  hence  (I.  15,  cor.  4)  they  are  all  squares.  Of 
these  squares,  also,  there  are  evidently  as  many  columns  as  there 
are  parts  in  AB  ;  while  in  each  column  there  are  as  many  squares 
as  there  are  parts  in  AD.  The  number  of  such  squares  con- 
tained in  a  figure  is  called,  in  the  language  of  mensuration,  the 
area  of  that  figure. 

Cor.  5.  Hence,  since  any  parallelogram  is  equivalent  (I.  15, 
cor.  5)  to  a  rectangle  on  the  same  base  and  between  the  same 


BOOK    I.]  EUCLID    AND    LEGENDKE.  41 

parallels,  it  follows  that  the  area  of  any  2^'^^'>'(Melogram  is 
equivale7it  to  the  product  of  its  base  and  its  "perpendicidar 
height  /  and  the  area  of  a  square  is  computed  by  multiplying 
a  side  by  itself 

Cor.  6.  Hence,  also  (I.  15,  cor.  4),  the  area  of  a  triangle  is 
computed  by  m.idtiplying  any  of  its  sides  by  the  perpendicidar 
draion  to  that  side  from  the  opposite  angle.,  and  taking  half 
the  product ;  and  the  area  of  a  trapezium  is  found  by  multi- 
P'lying  either  diagonal  by  the  sum  of  the  perpendicidar s  draion 
to  it  from  the  ayigles  which  it  subtends,  and  taking  half  the 
product.  When,  in  consequence  of  one  of  the  angles  being  re- 
entrant, the  perpendiculars  lie  on  the  same  side  of  the  diagonal, 
the  difference  of  the  perpendiculars  must  evidently  be  used  in- 
stead of  their  sum. 

Cor.  7.  Every  polygon  may  be  divided  into  triangles  or  tra- 
peziums by  drawing  diagonals ;  and  therefore  the  area  of  any 
polygon  whatever  can  be  computed  by  finding  the  areas  of 
those  component  figures  by  the  last  corollary,  and  adding  them 
together. 

Scho.  This  corollary  and  the  two  foregoing  contain  the  ele- 
mentary principles  of  the  mensuration  of  rectilineal  figures,  and 
they  form  a  connection  between  arithmetic  or  algebra  and  ge- 
ometry. They  also  explain  the  origin  of  the  expressions,  "  the 
square  of  a  number,"  "  the  rectangle  of  two  numbers,"  and  "  the 
product  of  two  lines." 

Prop.  XXIV. — Theor. — If  parallelograms  be  described  on 
two  sides  of  any  triangle,  and  their  sides  which  are  parallel  to  the 
sides  of  the  triangle  be  produced  until  they  meet,  the  sum  of 
the  parallelograms  will  be  equivalent  to  the  parallelogram  de- 
scribed on  the  base  of  the  triangle  having  its  adjacent  sides  to 
the  base  parallel  to  the  straight  line  joining  the  vertex  of  the 
triangle  with  the  point  of  intersection  of  the  sides  of  the  other 
parallelograms  produced,  and  terminated  by  the  latter  sides  or 
those  sides  produced. 

Let  BAG  be  a  triangle ;  the  parallelograms  MB  AD  and  CEFA, 
described  on  the  two  sides  BA  and  CA,  respectively,  are  to- 
gether equivalent  to  the  parallelogram  BCHK,  described  on 
the  base  BC  ;  the  parallelograms  MBAD,  CEFA,  and  BCIIK 


43 


THE   ELEMENTS   OF 


[book  I. 


beino;  described  ao;reeablv  to  the 
proposition. 

Describe  on  BA  and  CA  (L 
23)   tlie  parallelograms   MB  AD 
and   CEFA ;    and  produce   the 
M  B  L        c  E   titles   MD   and   EF   until    they 

meet  in  G ;  draw  GA,  and  produce  it  to  L  on  the  base  BC  (I. 
post.  2)  ;  describe  the  parallelogram  BCIIK  (I.  23)  on  BC. 
Then,  since  (const,  and  I.  15,  cor.  1)  BH  and  CK  are  parallel 
and  equal  to  AG,  they  are  parallel  and  equal  to  one  another  (I. 
ax.  1) ;  also  (1. 15,  cor  3)  HK  is  parallel  and  equal  to  BC ;  hence  (I. 
def.  15)  BCHK  is  a  parallelogram,  and  BLIIW  and  LCWK  are 
also  parallelograms.  Xow,  the  parallelograms  BLHW  and 
HBAG  (I.  15,  cor.  5)  are  equal,  and  for  similar  reason  the  parallel- 
ograms HBAG  and  MBDA ;  hence  (I.  ax.  1)  MBDA  is  equal  to 
BLHW.  In  similar  manner,  it  can  be  shown  tlnit  CEFA  is 
equiA^alent  to  LCKW;  therefore  the  whole  parallelogram  BCHK 
(I,  axs.  2  and  10)  is  equivalent  to  the  sum  of  the  two  parallelo- 
grams MB  AD  and  CEFA.  Wherefore,  if  on  any  two  sides  of 
a  triangle,  etc. 

Cor.  1.  A  paiticular  case  of  this  proposition  is,  lohen  the  tri- 
angle is  right-angled.,  then  the  sqiuires  described  on  the  legs — 
that  is,  the  sides  containing  the  right  angle,  are  together  equiva- 
lent  to  the  square  on  the  liypotlieniise — that  is,  the  side  opposite 
the  right  angle. 

Let  ABC  be  a  rio-ht-anoled 
triangle,  having  the  right  angle 
BAC ;  the  square  described  on 
tlie  hypothenuse  BC  is  equiva- 
lent to  the  sum  of  the  squares 
on  BA  and  AC.  On  BC,  BA, 
Q  and  AC  (L  23)  desciibe  the 
squares  BCHK,  BADE,  and 
ACFG ;  through  A  draw  AL 
parallel  to  BH  (L  18),  and 
draw  AH  and  DC  (L  post.  1). 
Tlien,  because  the  angles 
BAC  and  BAE  are  both  right 
angles    (L    def    10),   the    two 


BOOK  I.] 


EUCLID  AND  LEGENDEE. 


43 


straight  lines  CA  and  AE  are  in  the  same  straight  line  (I,  10). 
For  like  reason,  BA  and  AF  are  in  the  same  straight  line. 
Again  :  because  the  angle  IIBC  is  equal  to  the  angle  DBA  (I, 
ax.  1),  if  the  angle  ABC  be  added  to  each,  we  have  (I.  ax.  2) 
HBA  equal  to  DBC  ;  and  because  AB  is  equal  to  DB  (const.), 
and  BIT  equal  to  BC,  therefore  (I.  3,  case  1)  the  triangle  ABH 
is  equal  to  the  triangle  DBC.  But  (I.  15,  cor.  4)  the  parallelo- 
gram BPIIL  is  double  the  ti-iangle  ABH ;  for  like  reason  the 
square  BADE  is  double  the  triangle  DBC  ;  hence  (I.  ax.  6) 
BPIIL  is  equivalent  to  BADE.  In  like  manner,  PCLK  can  be 
shown  equivalent  to  ACFE.  Xow  (I.  ax.  10),  BCHK  is  equiva- 
lent to  BPHL  and  PCLK  together;  hence  (L  ax.  1)  BCHK  is 
equivalent  to  BADE  and  ACFG  together.  Wherefore,  if  the 
triangle  is  rio-ht-angled,  etc. 


OTHERWISE, 

Let  the  squares  on  AB  and 
BC  fall  on  the  same  side  of  BC. 
Describe  the  square  BAED  on 
the  side  BA  (L  23),  and  th.e 
square  BCMN"  on  the  side  BC 
(L  23),  and  produce  AE  to  F 
(I.  post.  2) ;  then  tlirough  D  draw 
PF  parallel  to  BC  (I.  18). 

Because  AF  is  parallel  to  BD 
(L  def.  14,  and  15,  cor.  1),  BC  is 
equal  to  DF,  and  BA  is  equal  to  '^^ 
DE,  and  the  angles  BAC  and  DEF  are  both  right  angles  (1. 
def  10),  and  equal  (I.  ax,  11);  therefore  the  triangle  BAC  is 
equal  to  the  triangle  DEF  (I.  3)  ;  and  because  BCED  is  com- 
mon to  the  square  BAED  and  the  parallelogram  (I.  def  14) 
BCDF,  and  the  triangles  BAC  and  DEF  equal,  the  square 
BAED  is  equivalent  to  the  parallelogram  BCDF.  And  PF 
being  parallel  to  BC  (const.),  the  parallelograms  (I.  def  14) 
BCDF  and  BCPL  have  a  common  base,  BC,  and  equal  altitudes; 
hence  (I.  15,  cor.  5)  they  are  equivalent,  and  (I.  ax,  1)  the 
square  BAED  is  equivalent  to  the  parallelogram  BCPL.  From 
A  draw  (I.  18)  AK  parallel  to  BM,  and  produce  DE  (I.  post. 
2)  to  BM;  then  AK  and  BM  being  parallel  (const.),  ED  and 


44 


THE   ELEMENTS    OF 


[book  I. 


BA,  being  opposite  sides  of  the  same  square,  are  also  parallel 
(I.  def.  15) ;  hence  MR  is  equal  to  BA,  and  equal  also  to  DE 
(I.  ax.  1).  But  BM  is  equal  to  BC  (const.) ;  therefore  the  par- 
allelogram BAjMR  is  equal  to  the  parallelogram  BCDF ;  and 
BAMR  having  the  same  base  and  equal  altitude  with  the  par- 
allelogram BGMK,  is  equivalent  to  it  (I.  15,  cor.  5);  hence 
BGMK  is  equivalent  to  BCDF,  equivalent  to  BCPL,  aud 
equivalent  to  the  square  BADE  (I.  ax.  1). 

Or,  the  square  described  upon  BA  is  equivalent  to  the  rect- 
angle of  the  hypothenuse  B  C  and  the  part  BG  of  the  hypjoth- 
enuse  nearest  to  BA  intercepted  by  the  p)erpe7idicular  drawn 
from  the  vertex  of  the  right  angle  to  the  liypothenuse.  In  a 
similar  manner,  it  can  be  shown  that  the  square  described  on 
A  G  is  equivalent  to  the  rectangle  of  the  hypothenuse  B  (7,  and 
the  remaining  part  G  G  of  the  hypothenuse  intercepted  by  the 
same  p)erpendicxdar.  But  the  two  rectangles  are  equivalent  to 
the  square  of  the  hyi^othenuse  (I.  ax.  10) ;  hence  the  tico  squares 
described  on  the  sides  AB  and  AG  (I.  ax.  1)  are  equivalent  to 
the  square  described  on  the  hypothenuse. 

E  F  And  if  we  make  the  tiiangle 

an  isosceles  rio;ht-an(2;led  tri- 
angle  as  ABC,  the  square  de- 
sciibed  on  AB  will  contain 
four  equal  triangles,  ACB,BCF, 
FCE,  and  ECA,  while  each  of 
the  squares  described  on  AC 
and  CB  will  contain  two  such  triangles,  and  both  together  will 
be  equivalent  to  the  four  equal  triangles,  or  equivalent  to  the 
square  on  AB.  The  demonstration  is  very  simple,  and  it  would 
be  well  for  the  pupil  to  undertake  it. 

Hence,  conversely,  if  the  square  described  upon  one  side  of  a 
triangle  be  equivalent  to  the  sum  of  the  square  described  upon 
the  two  other  sides  of  the  triangle,  the  angle  contained  by  those 
two  sides  is  a  right  angle ;  and  when  those  two  sides  form  two 
equal  squares,  the  triangle  is  a  right-angled  isosceles  triangle. 

Scho.  1.  The  proof  of  the  corollary  can  be  shown,  also,  either 
by  describing  the  square  of  the  hypothenuse  on  the  other  side 
of  BC;  and  the  other  squares  sometimes  on  one  side  and  some- 
times on  the  other;  and  since  drawing  a  perpendicular  from  the 


BOOK   I.]  EUCLID   AND    LEGEXDKE.  45 

vertex  of  tlie  riglit  angle  to  the  hypothenuse  makes  two  ri(iht 
angles,  so  a  line  can  be  drawn  from  the  same  vertex  to  the 
point  of  bisection  of  the  hypothenuse  and  make  two  supple- 
niental  avgles  and  two  equivalent  triangles,  and  the  demonsti'a- 
tion  conducted  by  supplemental  angles  (I.  15,  cor.  V)  instead  of 
right  angles.  Proportion  also  gives  neat  and  easy  solutions  to 
this  corollary.     (See  V.  8,  scho.) 

Cor.  2.  If  two  right-angled  triangles  have  their  hypothenuses 
equal,  and  a  side  similarly  situated  in  each  also  equal,  the  two 
triangles  are  equal  by  the  third  proposition  of  this  book ;  and, 
conversely,  if  the  legs  of  a  right-angled  triangle  be  equal  to  the 
legs  of  another  right-angled  triangle,  each  to  each,  their  hypoth- 
enuses can  be  in  a  similar  manner  shown  equal. 

Cor.  3.  Hence,  also,  we  can  find  a  square  equivalent  to  the 
sum  of  more  than  two  squares  ;  thus,  let  AB  be  the  side  of  one 
square,  and  AC,  perpendicular  to  it,  the  side  of  another  squai'e ; 
join  CB ;  the  square  on  CB  (I,  24,  cor.  l) 
is  equivalent  to  the  sum  of  the  squares  on 
CA  and  AB.  In  like  manner,  if  CD  be 
drawn  perpendicular  to  CB,  and  DB  be 
drawn,  the  square  on  DB  is  equivalent  to 
the  squares  on  DC,  CA,  and  AB,  and  by 
drawing  a  perpendicular  to  DB,  a  square  can  be  found  equiva- 
lent to  the  sum  of  four  squares ;  hence  a  square  can  be  found 
equivalent  to  the  sum  of  any  number  of  squares. 

Cur.  4.  Since  (CB)-  o  (AB)=  +  (CA)=,  we  have  (AB)=o 
(CB)"  —  (CA)-;  hence  a  square  can  be  found  equivalent  to  the 
difference  of  two  squares. 

Cor.  5.  If  a  perpendicular  be  drawn  from  the  vertex  of  the 
angle  A,  in  the  triangle  BAC  (diagram  to  cor.  l),  to  P  on  the 
hypothenuse  BC,  cutting  BC  into  two  segments,  BP  and  PC, 
the  difference  of  the  squares  on  the  sides  AB  and  AC  is  equiva- 
lent to  the  difference  of  the  squares  on  the  segments  BP  and 
PC.  For  the  square  on  AB  is  equivalent  to  the  squares  on  BP 
and  PA,  and  the  square  on  AC  is  equivalent  to  the  squares  on 
PC  and  PA;  therefore  (AC)^  —  (AB)^  =o-  (PC)'  —  (BP)^ 

Cor.  6.  Hence  the  squares  on  txoo  sides  of  a  triangle  are  to- 
gether equivalent  to  txoice  the  square  of  half  the  remaining  side^ 
and  twice  the  square  of  the  straight  line  from  its  point  of  M- 


46 


THE   ELEMENTS    OF 


[book  I. 


section  to  the  opposite  anrjle.  Suppose  P  in  tlic  triangle  to  te 
the  point  of  bisection  of  the  side  BC ;  tlion,  when  AP  is  per- 
pendicuhir  to  BC,  we  have  (AB)-  +  (AC)=  o  (BP)-  +  (AP)' 
+  (PC)'  +  (AP)^  .0=  2  (BP)=  +  2  (AP)\  And  when  AP  is 
not  perpendicular  to  BC,  the  equivalence  of  the  squares  on  two 
sides  to  the  same  will  be  shown  in  tlic  next  book. 

Scho.  2.  In  proof  of  coj-.  5,  the  obvious  principle  is  employed, 
that  the  difference  of  two  magnitudes  is  the  same  as  the  differ- 
ence obtained  after  adding  to  each  the  same  third  magnitude. 
Thus  the  difference  of  the  squares  on  BP  and  PC  is  the  same  as 
the  difference  between  the  sum  of  the  squares  BP  and  PA  and 
of  PC  and  PA. 


Pkop.  XXy. — Theok. —  Tlie  side  and  diagonal  of  a  square 
are  incommeiisuraUe  to  one  another — that  is,  there  is  no  Ivie 
which  is  a  measure  of  both. 

Let  ABCD  be  a  square,  and  BD  one  of  its  diagonals ;  AB, 
BD  are  incommensurable. 

Cut  off  DE  equal  to  DA,  and  join  AE.     Then,  since  (I.  1, 

cor.  1)  the  angle  DEA  is  equal  to  the  acute  angle  DAE,  AEB 

A  B     i^  obtuse,  and  therefore   (I.   22,  cor.)  in 

the  triangle  ABE,  BE  is  less  than  AB, 
or  than  AD  ;  wherefore  AD  is  not  a 
measure  of  BD.  Draw  EF  ])erpendicular 
to  BD.  Then  tiie  angles  FAE,  FEA, 
being  the  complements  of  the  equal 
angles  DAE,  DEA,  are  equal,  and 
therefore  AF,  FE  are  equal.  But  (I. 
20,  cor.  4)  ABD  is  half  a  riglit  angle ; 
as  is  also  BFE,  since  BEF  is  a  right  angle ;  wherefore  BE  is 
equal  to  FE,  and  therefore  to  AF.  From  FB,  which  is  evi- 
dently the  diagonal  of  a  square  of  which  FE  or  EB  is  the  side, 
cut  off  FG  equal  to  FE,  and  join  GE.  Then  it  would  be  shown, 
as  before,  that  BG  is  less  than  BE ;  and  therefore  BE,  the  dif- 
ference between  the  side  and  diagonal  of  tlie  square  .VC,  is  con- 
tained twice  in  the  side  AB,  with  the  remainder  GB,  wliich  is 
itself  the  difference  between  the  side  FE  or  EB,  and  the  diag- 
onal FB  of  another  squai*e.  By  repeating  the  process,  we 
Bhould  find,  in  exactly  the  same  manner,  that  BG  would  be 


BOOK   I.]  EUCLID    A:SD   LEGENDEE.  47 

contained  twice  in  BE,  with  a  remainder,  which  would  be  the 
difierence  between  the  side  and  diagonal  of  a  square  described 
on  I3G ;  and  it  is  evident  that  a  like  process  might  be  repeated 
continually,  as  no  excess  of  a  diagonal  above  a  side  would 
be  contained  in  the  side  witliout  remainder;  and  as  this  pro- 
cess has  no  termination,  theiv  is  no  line,  however  small,  which 
will  be  contained  without  remainder  in  both  AB  and  BD ;  they 
are,  therefore,  incommensurable. 

Scho.  Tills  proposition  can  be  illustrated  by  numbers.  Let 
10  be  the  side  of  tlie  square  ;  then  (I.  24,  cor.  1)  the  diagonal 
will  be  expressed  by  the  square  root  of  200,  or  14'142  +  ;  there 
being  no  common  multiple  of  10  and  14-142  +,  these  numbers 
are  incommensurable  with  each  otlier.  Or,  wlicn  any  two  lines 
are  taken  which  by  division  and  subdivision  no  common  meas- 
ure can  be  found  which  can  be  contained  in  each  v.'ithout  a  re- 
mainder, the  two  lines  are  said  to  be  ijicommensicrable  with 
each  other,  and  such  lines  are  tlie  side  and  diagonal  of  a  square  ; 
and  any  two  magnitudes  wiiatever  which  have  no  common 
unit  of  measure  are  incommensurable  with  one  another. 


E^TD    or   BOOK   FIRST. 


BOOK    SECOND.* 
ON"   THE   llECTANGLE   AND    SQUARE. 

DEFINITIONS. 

1.  A  rectangle  is  said  to  be  contained  by  the  two  straight 
lines  which  are  about  any  of  the  right  angles.  For  the  sake  of 
brevity,  the  rectangle  contained,  by  AB  and  CD  is  often  ex- 
pressed simply  by  AB.  CD,  a  jooint  being  placed  between  the 
letters  denoting  the  sides  of  tlie  rectangle ;  and  the  square  of  a 
line  AB  is  often  written  simply  AB". 

2.  A  gno7non  is  the  part  of  a  parallelogram  which  remains 
when  either  of  the  parallelograms  about  one  of  the  diagonals 
is  taken  away. 

PROPOSITIONS. 

Pkop.  I. — TiiEOR. — If  there  be  tioo  straight  lines,  one  of 
which  is  divided  into  any  number  of  parts,  the  rectangle  con- 
tained by  the  two  lines  is  equivalent  to  the  rectangles  contained 
by  the  undivided  line,  and  the  several  j^arts  of  the  divided  line. 

Let  A  and  BC  be  two  straight  lines;  and  let  BC  be  divided 
into  any  parts  in  tlie  points  D,  E  ;  tlie  rectangle  contained  by 
A  and  BC  is  equivalent  to  the  rectangles  contained  by  A  and 
BD,  A  and  DE,  and  A  and  EC. 

From  B  draAV  (I.  V)  BG  perpendicular  to  BC,  and  make  it 
equal  to  A ;  through  G  draw  (I.  1 8)  GH  par- 
^         DEC      ^^^^^  ^^  J3Q .  ^^^^  through  D,  E,  C  draw  DK, 

EL,  CH  parallel  to  BG.  Then  BH,  BK,  DL, 
and  EH  are  evidently  rectangles ;  and  BH  is 
equivalent  (I.  ax.  10)  to  BK,  DL,  EH.  But 
BH  is  contained  by  A  and  BC,  for  (H.  dcf.  1) 
■^  it  is   contained  by  GB   and  BC,  and   GB  is 

*  The  second  and  third  books  are  arrantred  very  similarly  to  those 
books  in  the  edition  of  Euclid  i)y  Professor  Thomson  of  the  University 
of  Glasgow,  Scotland. 


G  K        L  n 


BOOK  11.^  EUCLID  AND  LEGENDEE.  49 

equal  (const.)  to  A;  and  BK  is  contained  by  A  and  BD,  for  it 
is  contained  by  GB  and  BD,  of  whicli  GB  is  equal  to  A.  Also 
DL  is  contained  by  A  and  DE,  because  DK,  that  is  (I.  15,  con 
4)  BG,  is  equal  to  A ;  and  in  like  manner  it  is  shown  that  EH 
is  contained  by  A  and  EC.  Therefore  the  rectangle  contained 
by  A  and  BC  is  eqidvalent  to  the  several  rectangles  contained 
by  A  and  BD,  by  A  and  DE,  and  by  A  and  EC.  Wherefore, 
if  there  be  two  straight  lines,  etc. 

Prop.  II. — Theoe. — If  a  straight  line  he  divided  into  any 
ttoo  parts,  the  rectangles  contained  by  thexchole  and  each  of  the 
parts  are  together  equivalent  to  the  square  of  the  whole  line. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C  ;  the  rectangles  AB.AC  and  AB.BC  are  equivalent  tc» 
the  square  of  AB, 

Upon  AB  describe  (I.  23)  the  square  AE,  and  through  C 
draw  (I.  18)  CF  parallel  to  AD  or  BE.  Then 
AE  is  equal  (I.  ax.  10)  to  the  rectangles  AF  and 
CE.  But  iVE  is  the  square  of  AB,  and  AF  is 
the  rectangle  contained  by  BA,  AC ;  for  (II, 
def  1)  it  is  contained  by  DA,  AC,  of  which  DA 
is  equal  to  AB ;  and  CE  is  contained  by  AB, 
BC,  for  BE  is  equivalent  to  AB ;  therefore  the 
rectangles  under  AB,  AC,  and  AB,  BC  are  equivalent  to  the- 
square  of  AB.     If,  therefore,  etc. 

8cho.  This  proposition  may  also  be  demonstrated  in  the  fol- 
lowing manner : 

Take  a  straight  line  D  equal  to  AB,     Then  (II,  1)  the  rect- 
angles AC.D  and  BCD  are  together  equiva- 
lent to  AB.D.     But  since  D  is  equal  to  AB,     ^ ^    ^ 

the  rectangle  AB.D  is  equivalent  (I.  def  15) 

to  the   square   of  AB,   and   the   rectangles 

AC.D  and  BCD  are  respectively  equivalent  ° 

to  ACAB  and  BCAB ;  wherefoi-e  the  rect- 
angles ACAB   and  BCAB  are  together  equivalent   to    the 
square  of  AB. 

In  a  manner  similar  to  this,  several  of  the  following  proposi- 
tions may  be  demonstrated.     Such  proofs,  thougli  perhaps  not 
60  easily  understood  at  first  by  the  learner,  are  shorter  than 
4 


so  THE   ELEMENTS    OF  [bOOK   H. 

those  given  by  Euclid;  aud  they  have  tlie  advantage  of  being 
derived  from  those  preceding  them,  instead  of  being  estab- 
lished by  continual  appeals  to  original  principles. 

Pkop.  III. — Theor. — If  a  straight  line  he  divided  into  any 
two  2ici7'ts,  the  rectangle  contained  hy  the  lohole  and  one  of.  the 
parts  is  equivalent  to  the  square  of  that  part^  together  loith  the 
rectangle  contained  hy  the  tioo  parts. 

Let  the  straight  line  AB  be  divided  into  two  parts  in  the 
point  C  ;  the  rectangle  AB.BC  is  equivalent  to  the  square  of 
-BC,  together  with  the  rectangle  AC.CB 

Upon  BC  describe  (I.  23)  the  square  CE  ;  produce  ED  toF; 

and  through  A  draw  (I.  18)  AF  parallel 
to  CD  or  BE.  Then  the  rectangle  AE 
is  equivalent  (I.  ax.  10)  to  the  rectangles 
CE,  AD.  But  AE  is  the  rectangle  con- 
tained by  AB,  BC,  for  it  is  contained 
by  x^B,  BE,  of  which  BE  is  equal  to 
BC ;  and  AD  is  contained  by  AC,  CB, 
for  CD  is  equal  to  CB  ;  also  DB  is  the  square  of  BC.  Thei-e- 
fore  the  rectangle  AB.BC  is  equivalent  to  the  square  of  BC, 
together  with  the  rectangle  AC.CB.     If,  therefore,  etc. 

8cho.  Otherwise  :  Take  a  line  D  equal  to  CB.     Then  (II.  1) 
the   rectangle  AB.D    is    equivalent    to  the 

-^       ^ ^    rectangles  BCD  and  AC.D  ;  that  is  (const. 

and   L    def    15),   the   rectangle   AB.BC   is 
^  equivalent  to  the  square  of  BC  togctlier  with 

the  rectangle  AC.CB. 

Prop.  IV.— Tiieor.— 7/"  a  straight  line  he  divided  into 
any  tico  j^^-^^ts,  the  square  of  the  v-hole  line  is  equivalent  to 
the  squares  of  the  tioo  i)arts\  together  with  twice  their  red 

angle. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  C ; 
the  square  of  AB  is  equivalent  to  the  squares  of  AC  and  CB, 
together  with  twice  the  rectangle  under  AC  and  CB. 

On  AB  describe  (L  23)  the  square  of  AE,  and  join  BD; 
through  C  draw  (I.  18)  CGF  parallel  to  AD  or  BE;  and 
through  G  draw  UK  parallel  to  AB  or  DE.     Then,  l)ecause 


G 

/ 

BOOK  II.]  EUCLID  AND  LEGENDEE.  51 

CF  is  parallel  to  AD,  aucl  BD  falls  upon  them,  the  exterior 
angle  CGB  is  equivalent  (I.   16)  to  the  inte- 
rior and  remote  angle  ADB  ;  but  ADB  is         ^         ^ ? 

•equal  (I.  l)  to  ABD,  because  BA  and  AD 
are  equal,  being  sides  of  a  square;  wherefore 
(I.  ax.  1)  the  angle  CGB  is  equal  to  GBC; 
iiad  therefore  the  side  BC  is  equal  (I.  1,  cor.) 
to  the  side  CG.     But  (const.)  the  figure  CK         »  ^      B 

is  a  parallelogram  ;  and  since  CBK  is  a  right 
angle,  and  BC  equal  to  CG,  CK  (I.  def  15)  is  a  square,  and  it 
is  upon  the  side  CB.  For  the  like  reason  HF  also  is  a  square, 
and  it  is  upon  the  side  HG,  which  is  equal  (I.  15,  cor.  1)  to  AC; 
therefore  HF,  CK  are  the  squares  of  AC,  CB.  And  because 
(I.  15,  cor.  8)  the  complements  AG,  GEare  equivalent,  and  that 
AG  is  the  rectangle  contained  by  AC,  CB,  for  CG  has  been 
proved  to  be  equal  to  CB ;  therefore  GE  is  also  equivalent  to 
the  rectangle  AC.CB;  wherefore  AG,  GE  are  equivalent  to 
twice  the  rectangle  AC.CB.  The  four  figures,  therefore,  HF, 
-CK,  AG,  GE  are  equivalent  to  the  squares  of  AC,  CB,  and 
twice  the  rectangle  AC.CB.  But  HF,  CK,  AG,  GE  make  up 
the  Avhole  figure  AE,  which  is  the  square  of  AB ;  therefore  the 
square  of  AB  is  equivalent  to  the  squares  of  AC  and  CB,  and 
twice  the  rectangle  AC.CB.     Wherefore,  etc. 

Otherwise:  AB^  =  AB.AC  +  AB.CB  (H.  2).  But  (H.  3) 
A.B. AC =AC=+ AC.CB,  and  AB.CB=zCB-+AC.BC.  Hence 
<I.  ax.  2)  AB.AC  +  AB.BC,  or  AB-=AC^+CB-+2AC.CB. 

Cor.  1.  It  follows  from  this  demonstration,  that  the  parallel- 
ograms about  the  diagonal  of  a  square  are  likewise  squares. 

Cor.  2.  Hence  the  square  of  a  straight  line  •  is  equivalent  to 
four  times  the  square  of  its  half;  for  the  straight  line  being 
iDisected,  the  rectangle  of  the  parts  is  equivalent  to  the  square 
of  one  of  them. 

Pkop.  V. — Theor. — If  a  straight  line  be  divided  into  two 
'  equal  parts,  and  also  into  tioo  unequal  parts ;  the  rectangle  con- 
tained by  the  unequal  parts.,  together  with  the  square  of  the  line 
hettoeen  the  points  of  section,  is  equivalent  to  the  square  of  half 
the  line. 

Let  the  straio;ht  line  AB  be  bisected  in  C,  and  divided  un- 


H 

/ 

K                L 

/ 

52  THE    ELEMENTS    OF  [bOOK   II. 

equally  in  D  ;  the  rectangle  AD.DB,  together  with  the  square 
of  CD,  is  equivalent  to  the  square  of  CB. 

On  CB  describe  (I.  23)  the  square  CF,  join  BE,  and  through 
D  draw  (I.  18)  DHG  parallel  to  CE  or  BF ;  also  through  H 

draw  KLM  parallel  to  CB  or  EF  ;  and 
■^  c       D      B      through  A  draw  AK  parallel  to  CL  or 

BM.     Then   (I.   15,  cor.  5)  AL  and  CM 
^^    are  equal,  because  AC  is  equal  to  CB  ; 

and  (I.  15,  cor.  8)  the  complements  CH 
E       G      F       and   HF  are   equivalent.     Therefore    (I. 

ax.  2)  AL  and  CH  together  are  equal  to 
CM  and  HF  together ;  that  is,  AH  is  equivalent  to  the  gnomon 
CMC  To  each  of  those  add  LG,  and  (I.  ax.  2)  the  gnomon 
CMG,  together  with  LG,  is  equivalent  to  AH  together  with 
LG.  But  the  gnomon  CMG,  and  LG  make  up  the  figure 
CEFB,  which  is  the  square  of  CB  ;  also  AH  is  the  rectangle 
imder  AD  and  DB,  because  DB  is  equal  (IL  4,  cor.  1)  to  DH  ; 
and  LG  is  the  square  of  CD.  Thei-efore  the  rectangle  AD.DB 
and  the  square  of  CD  are  equivalent  to  the  square  of  CB. 
Wherefore,  if  a  straight  line,  etc. 

Otherwise  :  Since,  as  is  easily  seen  from  the  proof  in  the 
above,  DF  is  equal  to  AL,  take  these  separately  from  the  entire 
figure,  and  there  remain  AH  and  LG  equivalent  to  the  square 
CF,  as  before.     The  proof  may  also  be  as  follows: 

AD.DB  =  CD.DB  +  AC.DB  (H.  1)  or  AD.DB =CD.DB+ 
CB.DB,  because  CB=:AC.  Hence  (H.  3)  AD.DB =CD.DB + 
CD.DB+DB^=i:2CD.DB  +  DBl  To  each  of  these  add  CD=; 
then  AD.DB  +  CD^=2CD.DB+DB^+CD^  or  (H.  4)  AD.DB 
+  CD'=CB\ 

Cor.  1.  Hence  the  ditference  of  the  squares  of  CB  and  CD  is 
equivalent  to  the  rectangle  under  AD  and  DB.  But  since  AC 
is  equal  to  CB,  AD  is  equivalent  to  the  sum  of  CB  and  CD,  and 
DB  is  the  difterence  of  these  lines.  Hence  the  difference  of  the 
squares  of  two  straight  lines  is  equivalent  to  the  rectangle  under 
their  sum  and  difference. 

Cor.  2.  Since  the  square  of  CB,  or,  which  is  the  same,  the 
rectangle  AC.CB,  is  greater  than  the  rectangle  AD.DB  by  the 
equare  of  CD,  it  follows,  that  to  divide  a  straight  line  into  two 


BOOK  n.]  EUCLID  AND  LEGENDKE.  53 

parts,  the  rectangle  of  wliich  may  be  the  greatest  possible,  or, 
as  is  termed,  a  tnaxlmiim,  tlie  line  is  to  be  bisected. 

Cor.  3.  Hence  also  tlie  sum  of  the  squares  of  tlie  two  parts 
into  which  a  straight  line  is  divided,  is  the  least  possible,  or  is, 
as  it  is  termed,  a  ininimuni,  Avlien  the  line  is  bisected.  For  (II. 
4)  the  square  of  tlie  line  is  equivalent  to  the  squares  of  the 
parts  and  twice  their  rectangle  ;  and  therefore  the  greater  the 
rectangle  is,  the  less  are  the  squares  of  the  j^arts;  but,  by  the 
foregoins;  corollarv,  the  rectangle  is  a  maximum  when  the  line 
is  bisected. 

C<yr.  4.  Since  (I.  24,  cor.  5)  the  difference  of  the  squares  of 
the  sides  of  a  triangle  is  equivalent  to  the  difference  of  the 
squares  of  the  segments  of  the  base,  it  follows,  from  the  first 
corollary  above,  that  the  rectangle  under  the  sum  and  differ- 
ence of  the  sides  of  a  triangle  is  equivalent  to  the  rectano-le 
under  the  sum  and  difference  of  the  segments,  intercepted  be- 
tween the  extremities  of  the  base  and  the  jjoint  in  which  the 
perpendicular  cuts  the  base,  or  tlie  base  produced. 

Cor.  5.  Hence,  also,  if  a  straight  line  be  drawn  from  the  ver- 
tex of  an  isosceles  triangle  to  any  point  in  the  base,  or  its  con- 
tinuation, the  difference  of  the  squares  of  that  line  and  either  of 
the  equal  sides  is  equivalent  to  the  rectangle  under  the  seoments 
intercepted  between  the  extremities  of  the  base  and  the  jDoint. 

Cor.  6.  Since  (I.  24,  cor.  4)  the  square  of  one  of  the  legs  of  a 
right-angled  triangle  is  equivalent  to  the  difference  of  the 
squares  of  the  hypothenuse  and  the  other  leg,  it  follows  (IT.  5, 
cor.  1)  that  the  square  of  one  leg  of  a  right-angled  triangle  is 
equivalent  to  the  rectangle  under  the  sum  and  difference  of  the 
hypothenuse  and  the  other. 

^cho.  And  a  parallelogram  can  be  constructed  equivalent 
to  a  ffiven  triangle,  or  anv  given  rectilinear  fiorure  havinor  an 
angle  equal  to  a  given  angle,  or  applied  to  a  given  straight  line 
— that  is,  having  that  straight  line  for  one  of  its  sides,  when 
the  jjarallelogram  shall  be  equivalent  to  a  given  triangle  or 
given  rectilinear  figure,  and  have  one  of  its  angles  equal  to  a 
given  angle,  by  applying  a  parallelogram  equivalent  to  the 
given  triangle  with  an  equal  angle,  to  a  given  straight  line,  and 
then  constructing  an  equal  triangle  to  the  given  triangle  (I.  15 
and  15,  cor.  4). 


A              C 

B     D 

K 

H 

/ 

L 

/ 

54  THE   ELEMENTS    OF  [bOOK    II, 

Prop.  VI. — Theok. — If  a  straight  line  he  bisected,  and  be 
produced  to  any  point,  the  rectangle  contained  by  the  whole 
line  thus  produced,  and  the  part  of  it  produced,  together  with 
the  square  of  half  the  line  bisected,  is  equivalent  to  the  square 
of  the  straight  line  which  is  made  %ip  of  the  half  and  the  p>art 
produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to 
D ;  the  rectangle  AD.DB,  and  the  square  of  CB,  are  equiva- 
lent to  the  square  of  CD. 

Upon  CD  describe  (I.  23)  the  square  CF,  and  join  DE ; 
through  B  draw  (I.  18)  BHG  parallel  to  CE  or  DF ;  through 

H  draw  KLM  parallel  to  AD  or  EF,  and 
through  A  draw  AK  parallel  to  CL  or 
jj  DM.  Then  because  AC  is  equal  to  CB, 
the  rectangle  AL  is  equal  (I.  15,  cor.  5)  to 
CH ;  but  (1.15,  cor.  8)  the  complements  CH, 
J,  ^j    J,     HF  are  equivalent ;  therefore,  also,  AL  is 

equal  to  HF,  To  each  of  these  add  CM 
and  LG ;  therefore  AM  and  LG  are  equivalent  to  the  whole 
square  CEFD,  But  AM  is  the  rectangle  under  AD  and  DB, 
because  (11.  4,  cor.  1)  DB  is  equal  to  DM;  also,  LG  is  the 
square  of  CB,  and  CEFD  the  square  of  CD.  Therefore  the 
rectangle  AD.DB  and  the  square  of  CB  are  equivalent  to  the 
square  of  CD.     Wherefore,  if  a  straight  line,  etc. 

Otherwise  :  Produce  CA  to  N,  and  make  CN  equal  to  CD. 

To  these  add  the  equals  CB  and  CA  ; 
^  ^  ^  ^  °     therefore  NB  is  equal  (L  ax.  2)  to  AD. 

But  (IL  5)  the  rectangle  NB.BD,  oi- 
AD.BD,  together  with  the  square  of  CB,  is  equivalent  to  the 
square  of  CD ;  which  was  to  be  proved. 

Prop.  VIL — Theor. — If  a  straight  line  be  divided  into  a?2y 
two  parts,  the  squares  of  the  whole  line  and  of  one  of  the  p>arts 
are  equivalent  to  ticice  the  rectangle  contained  by  the  whole  and 
that  part,  together  tcith  the  square  of  the  other  part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C ;  the  squares  of  AB,  BC  are  equivalent  to  twice  the 
rectangle  AB.BC,  together  with  the  square  of  AC. 


BOOK  II.]  EUCLID  AND  LEGENDRE.  65 

Upon  AB  describe  (I.  23)  the  square  AE,  and  construct  the 
figure  as  in  the  preceding  propositions.  Then, 
because  (I.  15,  cor.  8)  the  complements  CII,  ^  c      B 

FK  are  equivalent,  add  to  each  of  thorn  CK; 
the  whole  AK  is  therefore  equal  to  the  whole       ^ 
CE ;  therefore  AK,  CE  are  together  double 
of  AK.     But  AK,  CE  are  the  gnomon  AKF, 


K 


together  with  the  square  CK  ;  therefore  the         i>  F      E 

gnomon  AKF  and  the  square  CK  are  double 
of  AK,  or  double  of  the  rectangle  AB.BC,  because  BC  is  equal 
(II.  4,  cor.  1)  to  BK.  To  each  of  these  equals  add  HF,  which 
is  equal  (II.  4,  cor.  1 )  to  the  square  of  AC  ;  therefore  the  gno- 
mon AKF,  and  the  squares  CK,  HF  are  equivalent  to  twice  the 
rectangle  AB.BC  and  the  square  of  AC.  But  the  gnomon 
AKF,  and  the  squares  CK,  HF  make  up  the  whole  figure  AE, 
together  with  CK;  and  these  are  the  squares  of  AB  and  BC ; 
therefore  the  squares  of  AB  and  BC  are  equivalent  to  twice  the 
rectangle  AB.BC,  together  with  the  square  of  AC ;  wherefore, 
if  a  straight  line,  etc. 

Otherwise:  Because  (II.  4)  AB^  =  AC=+BC-+2AC.BC,  add 
BC=  to  both;  then  AB=+BC==AC=  +  2BC=+2AC.BC.  But 
(II.  3)  BC=+AC.BC= AB.BC,  and  therefore  2BC=+2AC.BC 
=2AB.BC;  wherefore  AB-+BC^=AC=+2AB.BC. 

Cor.  1.  Since  AC  is  the  diflierence  of  AB  and  BC,  it  follows 
that  the  square  of  the  difterence  of  two  straight  lines  is  equiva- 
lent to  the  sum  of  their  squares,  wanting  twice  their  rectangle. 

Cor.  2.  Since  (II.  4)  the  square  of  the  sum  of  two  lines  ex- 
ceeds the  sum  of  their  squares  by  twice  their  rectangle,  and 
since,  by  the  foregoing  corollary,  the  square  of  their  difference 
is  less  than  the  sum  of  their  squares  by  twice  their  rectangle,  it 
follows  that  the  square  of  the  sum  of  two  lines  is  equivalent  to 
the  square  of  their  difference,  together  with  four  times  their 
rectangle. 

Pkop.  VIII. — TiiEOR. — If  a  straight  line  he  divided  into  two 
equal,  and  also  into  tvw  imequal  parts,  the  squares  of  the  un- 
equal parts  are  together  double  of  the  square  of  half  the  linCj 
and  of  the  square  of  the  line  hetioeen  the  points  of  section. 

Let  the  straight  line  AB  be  divided  equally  in  C,  and  un- 


56 


THE    ELICMENTS    OF 


[book  n. 


E 


equally  in  D  ;  the  square?  of  AD,  DB  are  together  double  of 
the  squares  of  AC,  CD. 

From  C  draAV  (I.  7)  CE  j)ei-|ie!Klicular  to  AB,  and  make  it 
equal  to  AC  or  CB,  and  join  EA,  EB  ;  through  D  draw  (I.  18) 
DF  jjarallel  to  CE,  and  tlirougli  V  draw  FG  parallel  to  AB  ; 
and  join  AF.  Then  because  (const.)  the  triangles  ACE,  BCE 
are  right-angled  and  isosceles,  tlie  angles  CAE,  AEC,   CEB, 

O  O  7  O  7  3  7 

EBC  are  (I.  20,  cor.  4)  each  half  a  riglit  angle.  So  also  (I.  16, 
part  2)  are  EEC,  BED,  because  FG  is  parallel  to  AB,  and  ED 
to  EC  ;  and  for  the  same  reason  EGF,  ADF  are  right  angles. 
The  angle  AEB  is  also  a  right  angle,  its  parts  being  each  half 
a  right  angle.     Hence  (I.  1)  EG  is  equal  to  GF  or  CD,  and  ED 

to  DB.  Again  (I.  24,  cor.  1) :  the  square 
of  AE  is  equivalent  to  the  squares  of  AC, 
CE,  or  to  tAvice  the  square  of  AC,  since 
AC  and  CE  are  equal.  In  like  manner, 
the  square  of  EF  is  equivalent  to  twice 
the  square  of  GF  or  CD.  Now  (I.  24, 
^  cor.  1),  the  squares  of  AD  and  DF,  or 
of  AD  and  DB,  are  equivalent  to  the  square  of  AF ;  and  the 
squares  of  AE,  EF,  that  is,  twice  the  square  of  AC  and  twice 
the  square  of  CD,  are  also  equivalent  to  the  square  of  AF; 
therefore  (I.  ax.  1)  the  squares  of  AD,  DB  are  equivalent  to 
twice  the  square  of  AC  and  twice  the  square  of  CD.  If,  there- 
fore, a  straight  line,  etc. 

Otherwise:  DB^+2BC.CD=BC=+CD^  (II.  7),  or  DB=+ 
2AC.CD=:AC=  +  CD^-;  also  (11.  4)  AD==:AC^+CD=+2AC. 
CD.  Add  these  equals  togethei-,  and  from  the  sums  take 
2AC.CD;  then  AD-+DB^=2AC^+2CD^ 


Prop.  IX. — ^Theor. — If  a  straight  line  he  bisected,  and  pro- 
duced to  any  point,  the  squares  of  the  whole  line  thus  produced, 
and  of  the  part  of  it  produced,  are  together  double  of  the  square 
of  half  the  lifie  bisected,  and  of  the  square  of  the  line  made  up 
of  the  half  and  the  part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to 
D ;  the  squares  of  AD,  DB  arc  double  of  the  squares  of  AC, 
CD. 

From  C  draw  (I.  7)  CE  perpendicular  to  AB ;  and  make  it 


BOOK   II.] 


EUCLID    AND   LEGENDRE. 


57 


A 


equal  to  AC  or  CB ;  join  AE,  EB,  and  tln-ough  E  and  D  draw 
(I.  1 8)  EF  parallel  to  AB,  and 
DF  parallel  to  CE.  Then  be- 
cause the  straiu'lit  line  EF 
meets  the  parallels  EC,  FD, 
the  ano-les  CEF,  EFD  are 
equivalent  (I.  9)  to  two  right 
anixles  :  and  therefore  the  an- 
gles  BEF,  EFD  are  less  than 

two  right  angles;  therefore  (I.  19)  EB,  FD  will  meet,  if  pro- 
duced toward  B,  D  ;  let  them  meet  in  G,  and  join  AG.  Then 
it  would  be  proved,  as  in  the  last  proposition,  that  the  angles 
CAE,  xVEC,  CEB,  EBC  are  each  half  a  right  angle,  and  AEB 
a  right  angle.  BDG  is  also  a  right  angle,  being  equal  (I.  16) 
to  ECB,  since  (const.)  EC,  EG  are  parallel ;  DGB,  DBG  are 
each  half  a  right  angle,  being  equal  (I.  10  and  I.  11)  to  CEB, 
CBE,  each  to  each ;  and  FEG  is  half  a  right  angle,  being  (I. 
16)  equal  to  CBE.  It  would  also  be  proved,  as  in  the  last 
proposition,  that  the  squai-e  of  AE  is  twice  the  square  of  AC, 
and  the  square  of  EG  twice  the  square  of  EF  or  CD.  Now 
(I.  24,  cor.  l),the  squares  of  AD,  DG,  or  of  AD,  DB,  are  equiv- 
alent to  tlie  sipiare  of  AG;  and  the  squares  of  AE,  EG,  or  twice 
the  square  of  AC  and  twice  the  square  of  CD,  are  also  equiva- 
lent to  the  square  of  AG.  Therefore  (I.  ax.  1)  the  squares  of 
AD,  DB  are  equivalent  to  twice  the  square  of  AC  and  twice 
the  square  of  CD.     If,  therefore,  a  straight  line,  etc. 

Otherwise:  Produce   CA,   making   CH   equal   to    CD.     To 
these  add  CB,  CA ;  therefore  IIB,  AD  are 
equal.     Then  (II.  8)  IIB--+BD^  or  AD=+ 
BD^  =  2CD=-f2AC-. 

&eho.  The  nine  foregoing  propositions  may  all  be  proved 
very  easily  by  means  of  algebra,  in  connection  with  the  princi- 
ples of  mensuration,  already  established  in  the  corollaries  to 
the  23d  proposition  of  the  first  book.  Thus,  to  pi'ove  the  fourth 
proposition,  let  AC=:a,  CB  =  5,  and,  consequently,  AB=a+5. 
Now,  the  area  of  the  square  described  on  AB  will  be  found  (I. 
23,  cor.  4)  by  multiplying  a-\-h  by  itself.  This  product  is 
found,  by  performing  tlie  actual  operation,  to  be  «'+2aJ+^'; 
an  expression,  the  first  and  third  parts  of  which  are,  by  the 


n   A 


B    D 


58  THE   ELEMENTS    OF  [bOOK    H. 

same  corollary,  the  areas  of  the  squares  of  AC  and  CB,  and  the 
second  is  twice  the  rectangle  of  those  lines. 

In  like  manner,  to  prove  the  eighth,  adopting  the  same  nota- 
tion, Ave  have  the  line  which  is  made  up  of  the  whole  and  CB 
=  a  +  2$y  and,  multiplying  this  by  itself,  we  get  for  the  area  ot 
the  square  of  that  line,  cr+iab+ib",  or  a'-\-4{a-]-b)b,  the 
first  part  of  which  is  the  area  of  the  square  of  AC,  and  the 
second  four  times  the  area  of  the  rectangle  under  AB  and  CB. 

It  will  be  a  useful  exercise  for  the  student  to  prove  the  other 
propositions  in  a  similar  manner.  He  will  also  find  it  easy  to 
investigate  various  other  relations  of  lines  and  their  parts  by 
means  of  algebra. 

All  the  properties  delivered  in  these  propositions  hold  also 
respecting  numbers,  if  products  be  substituted  for  rectangles. 
Thus,  7  being  equal  to  the  sum  of  5  and  2,  the  square,  or  sec- 
ond power  of  7,  is  equal  to  the  squares  of  5  and  2  and  twice 
their  product ;  that  is,  49  =  25  +  4+20. 

Pkop.  X. — Peob. — To  divide  a  given  straight  line  into  txoo 
parts,  so  that  the  rectangle  contained  by  the  \chole  and  one  of 
the  parts  may  be  equivalent  to  the  square  of  the  other  part. 

Let  AB  be  the  given  straight  line  ;  it  is  required  to  divide  it 
into  two  parts,  so  that  the  rectangle  under  the  whole  and  one 
of  the  parts  may  be  equivalent  to  the  square  of  the  other. 

Upon  AB  describe  (I.  23)  the  square  AD ;  bisect  (I.  6)  AC 
in  E,  and  join  E  with  B,  the  remote  extremity  of  AB  ;  produce 
CA  to  F,  making  EF  equal  to  EB,  and  cut  off  AH  equal  to  AF ; 
AB  is  divided  in  H,  so  that  the  rectangle  AB.BH  is  equivalent 
to  the  square  of  AH. 

Complete  the  parallelogram  AG,  and  produce  GH  to  K. 
Then,  since  BAC  is  a  right  angle,  FAH  is  also  (I.  9)  a  right 
angle;  and  (I.  def  15)  AG  is  a  square,  AF,  AH  being  equal 
by  construction.  Because  the  straight  line  AC  is  bisected  in 
E,  and  produced  to  F,  the  rectangle  CF.FA  and  the  square  of 
AE  are  together  equivalent  (II.  6)  to  the  square  of  EF  or  of  EB, 
since  (const.)  EB,  EF  are  equal.  But  the  squares  of  BA,  AE 
are  equivalent  (I.  24,  cor.  1)  to  the  square  of  EB,  because  the 
angle  EAB  is  a  right  angle;  therefore  the  rectangle  CF.FA 
and  the  square  of  AE  are  equivalent  (I.  ax.  1)  to  the  squares  of 


BOOK  n.] 


EUCLID    AND    LEGENDRE. 


59 


BA,  AE.     Take  away  the  square  of  AE,  -u-hich  is  common  to 
both;  therefore  the  remainino-  rectano-le  CF.FA 


F 


G 


E 


H 

-" 

B 


D 


is  equivalent  (I.  ax.  3)  to  the  square  of  AB. 

But  the  figure  FK  is  the  rectangle  contained 

by  CF,  FA,  for  AF  is  equal  to  FG  ;  and  AD 

is  the  square  of  AB  ;  therefore   FK  is  equal 

to  AD.     Take    away  the  common  part  AK, 

and  (I.  ax.  3)  the  remainders  FH  and  HD  are 

equivalent.     But  HD  is  the  rectangle  AB.BH, 

for  AB  is  equal  to  BD  ;  and  FH  is  the  square 

of  AH.     Therefore  the  rectangle  AB.BH  is  equivalent  to  the 

square  of  AH  ;  Avherefore  the  straight  line  AB  is  divided  in  H, 

so  that  the  rectangle  AB.BH  is  equal  to  the  square  of  AH ;, 

which  was  to  be  done. 

Sc/w.  1.  In  the  practical  construction  in  this  proposition,  and 
in  the  2d  cor.  to  9  of  the  fifth  book,  which  is  virtually  the  same,  it 
is  sufficient  to  draw  AE  perpendicular  to  AB,  making  it  equal 
to  the  half  of  AB,  and  producing  it  through  A ;  then,  to  make 
EF  equal  to  the  distance  from  E  to  B,  and  AH  equal  to  AF. 
It  is  plain  that  BD  might  be  bisected  instead  of  AC,  and  that 
in  this  way  another  point  of  section  would  be  obtained. 

While  the  enunciation  in  the  text  serves  for  ordinary  pur- 
poses, it  is  too  limited  in  a  geometrical  sense,  as  it  comprehends 
only  one  case,  excluding  another.  The  following  include& 
both : 

In  a  given  straight  line^  or  its  continuatio7i,  to  find  ai^ointy 
such  that  the  rectangle  contained  hy  the  given  line^  and  the  seg- 
ment between  one  of  its  extremities  and  the  required  point,  may 
be  equal  to  the  square  of  the  segment  between  its  other  extremity 
and  the  same  point. 

The  point  in  the  continuation  of  BA  will  be  found  by  cutting 
ofiT  a  line  on  EC  and  its  continuation,  equal  to  EB,  and  describ- 
ing on  the  line  composed  of  that  line  and  AE  a  square  lying  on 
the  opposite  side  of  AC  from  AD ;  as  the  angular  j^oint  of  that 
square  in  the  continuation  of  BA  is  the  point  required.  The 
proof  is  the  same  as  that  given  above,  except  that  a  rectangle 
corresponding  to  AK  is  to  be  added  instead  of  being  sub- 
tracted. 

Scho.  2.  The  line  CF  is  equivalent  to  BA  and  AH ;  and  since 


60  THE   ELEMENTS    OF  [bOOK   H, 

it  has  been  shown  that  the  rectangle  CF.FA  is  eqiiivalent  to 
the  square  of  BA  or  CA,  it  follows  that  if  any  straight  line  AB 
(see  the  next  diagram)  be  divided  according  to  this  proposition 
in  C,  AC  being  the  greater  part,  and  if  AD  be  made  equal  to 
AB,  DC  is  similarly  divided  in  A.     So  also  if  DE  be  made 


F 


E  D  A  C  B 


equal  to  DC,  and  EF  to  EA,  EA  is  divided  similarly  in  D,  and 
FD  in  E  ;  and  the  like  additions  may  be  continued  as  far  as  we 
please. 

Conversely,  if  any  straight  line  FD  be  divided  according  to 
this  proposition  in  E,  and  if  EA  be  made  equal  to  EF,  DC  to 
DE,  etc.,  EA  is  similarly  divided  in  D,  DC  in  A,  etc.  It  fol- 
lows also,  that  the  greater  segment  of  a  line  so  divided  will  be 
itself  similarly  divided,  if  a  part  be  cut  oif  from  it  equal  to  the 
less ;  and  that  by  adding  to  the  whole  line  its  greater  segment, 
another  line  will  be  obtained,  which  is  similarly  divided. 

Prop.  XI. — Theok. — In  an  oUuse-angkd  triangle^  the 
square  of  the  greatest  side  exceeds  the  squares  of  the  other  tioo^ 
hy  tioice  the  rectangle  contained  by  either  of  the  last-mentioned 
sides,  and  its  continuation  to  meet  a  perpendicular  drawn  to  it 
from  the  opposite  angle. 

Let  ABC  be  a  triangle,  having  the  angle  ACB  obtuse ;  and 
let  AD  be  perpendicular  to  BC  produced ;  the  square  of  AB  is 
equivalent  to  the  squares  of  AC  and  CB,  and  twice  the  rectan- 
gle BC.CD. 

Because  the  straight  line  BD  is  divided  into  two  parts  in  the 

point  C,  the  square  of  BD  is  equivalent  (II.  4)  to  the  squares 

of  BC   and   CD,  and  tAvice   the  rectangle 

BC.CD.     To  each  of  these  equivalents  add 

the  square  of  DA  ;  and  the  squares  of  DB, 

DA  are  equivalent  to  the  squares  of  BC, 

CD,  DA,  and  twice  the  rectangle  BC.CD. 

But,  because  the  angle  D  is  a  right  angle, 

the  square  of  BA  is  equivalent  (I.  24,  cor.  1)  to  the  squares  of 

BD,  DA,  and  tlie  square  of  CA  is  equivalent  to  the  squares  of 


BOOK  II.]  EUCLID  AND  LEGENDRE.  61 

CD,  DA ;  therefore  the  square  of  BA  is  equivalent  to  the 
squares  of  BC,  CA,  and  twice  the  rectangle  BC.CD.  Therefore, 
in  an  obtuse-angled  triangle,  etc. 

Pkop.  XII. — TiiEOR. — I/i  any  triangle^  the  square  of  a  side 
subtending  an  acute  a?igle  is  less  than  the  squares  of  the  other 
sides,  by  twice  the  rectangle  contained  by  either  of  those  sides, 
and  the  straight  line  hitercej^ted  between  the  acute  angle  and 
the  perpendicular  drawn  to  that  side  from  tJie  opposite  angle. 

Let  ABC  (see  this  figure  and  that  of  the  foregoing  proposi- 
tion) be  any  triangle,  having  the  angle  B  acute ;  and  let  AD 
be  perpendicular  to  BC,  one  of  the  sides  containing  that  angle; 
the  square  of  AC  is  less  than  the  squares  of  AB,  BC,  by  twice 
the  rectangle  CB.BD. 

The  squares  of  CB,  BD  are  equivalent  (II.  1)  to  twice  the 
rectangle  contained  by  CB,  BD,  and  the  square  of  DC.  To 
each  of  these  equals  add  the  square  of 
AD  ;  therefore  the  squares  of  CB,  BD, 
DA  are  equivalent  to  twice  the  rect- 
angle CB.BD,  and  the  squares  of  AD, 
DC.  But,  because  AD  is  perpendicu- 
lar to  BC,  the  square  of  AB  is  equiva- 
lent (I.   24,  cor.   1)  to  the   squares   of 

BD,  DA,  and  the  square  of  x\.C  to  the  squares  of  .VD,  DC  ; 
therefore  the  squares  of  CB,  BA  are  equivalent  to  the  square  of 
AC,  and  twice  the  rectangle  CB.BD ;  tliat  is,  tlie  square  of  AC 
alone  is  less  than  the  squares  of  CB,  BA,  by  twice  the  rectangle 
CB.BD. 

If  the  side  AC  be  perpendicular  to  BC,  then  BC  is  the 
straight  line  between  the  perpendicular  and  the  acute  angle  at 
B ;  and  it  is  manifest  that  the  squares  of  AB,  BC  are  equiva- 
lent (I.  24,  cor.  1)  to  the  square  of  AC  and  twice  the  square  of 
BC.     Therefore,  in  any  triangle,  etc. 

Scho.  By  means  of  this  or  the  foregoing  proposition,  the  area 
of  a  triangle  may  be  computed,  if  the  sides  be  given  in  num- 
bers. Thus,  let  AB  =  17,  BC  =  28,  and  AC  =  25.  From  AB=+ 
BC  take  AC;  that  is,  from  l7'4-2SUake  25=;  the  remainder 
448  is  twice  CB.BD.  Dividing  this  by  56,  twice  BC,  the  quo- 
tient 8   is   BD,      Hence,  from  either  of  the  triangles  ABD, 


62 


THE    ELEMENTS    OF  [bOOK   II. 


ACD,  we  find  the  perpendicular  AD  to  be  15  ;  and  thence  the 
area  is  found,  by  taking  half  tlie  product  of  BC  and  AD,  to  be 

210. 

The  segments  of  the  base  are  more  easily  found  by  means  -of 
tJie  4th  corollary  to  the  fifth  proposition  of  this  book,  in  connec- 
tion with  the  pi-inciple,  that  if  half  the  difference  of  two  mag- 
nitudes le  added  to  half  their  sum ^  the  resvlt  is  the  greater;  and 
if  half  the  difference  he  taken  from  half  the  sum,  the  remainder 
is  the  less.  Thus,  if  42,  the  sum  of  AB  and  AC,  be  multiplied 
Tjy  8,  their  difference,  and  if  the  product,  336,  be  divided  by  28, 
the  sum  of  the  segments  of  the  base,  the  quotient  12  is  their 
difference.  The  half  of  this  being  added  to  the  half  of  28,  the 
sum  20  is  the  greater  segment  CD ;  and  being  subtracted  from 
it,  the  remainder  8  is  BD. 

To  prove  the   principle    mentioned    alxjve,  let  AB    be   the 

greater,  and  BC  the  less  of  two 
^      ^  ^  ^       ^     magnitudes.     Bisect  AC  in  D,  and 

'  make  AE  equal  to  BC.     Then  AD 

or  DC  is  half  the  sum,  and  ED  or 
DB  half  the  difference  of  AB  and  BC  ;  and  AB  the  greater  is 
equivalent  to  the  sum  of  AD  and  DB,  and  BC  the  less  is  equiv- 
alent to  the  difference  of  DC  and  DB. 

Cor.  Hence,  when  AP  (I.  24,  cor.  6)  is  not  perpendicular  to 
BC — the  truth  of  the  corollary  can  be  shown  by  this  and  pre- 
-ceding  propositions. 

Hence,  if  the  sides  of  a  ti-iangle  be  given  in  numbers,  the  line 
AD  can  be  computed.  Thus,  if  AC^ll,  AB  =  14,  and  BC  =  7, 
we  have  AC-f  BC=  =  121  +  49  =  170,  and  2AD-=:98.  Then  170 
— 98  =  72,  the  half  of  Avhich  is  36  ;  and  6,  the  square  root  of 
this,  is  CD. 

Prop.  XHI. — Vrov,.— To  describe  a  square  that  shall  he 
•equivalent  to  a  given  rectilineal  figure. 

Let  A  be  the  given  rectilineal  figure ;  it  is  required  to  de- 
scribe a  square  that  shall  be  equal  to  it. 

Describe  (H.  5  scho.)  the  rectangle  BD  equal  to  A.  Then,  if  the 
sides  of  it,  BE,  ED,  be  equal  to  one  another,  it  is  a  square,  and 
what  was  required  is  done.  But  if  they  be  not  equal,  produce 
one  of  them  BE  to  F,  and  make  EF  equal  to  ED;  bisect  BF  ia 


BOOK  II.] 


EUCLID  AND  LEGENDEE. 


63 


O,  and  from  the  center  G,  at  the  distance  GB,  or  GF,  describe 
(I.  post.  3)  the  semicircle  BIIF; 
produce  DE  to  H,  and  join  GH. 
Therefore,  because  the  straight 
line  BF  is  divided  equally  in  G, 
and  unequally  in  E,  the  rectangle 
BE.EF,  and  the  square  of  EG, 
are   equivalent    (II.    5)    to    the 

square  of  GF,  or  of  GH,  because  GH  is  equal  to  GF.  But  the 
squares  of  HE,  EG  are  equal  (I.  24,  cor,  l)  to  the  square  of  GH ; 
therefore  the  rectangle  BE.EF  and  the  square  of  EG  are  equiv- 
alent to  the  squares  of  HE,  EG.  Take  away  the  square  of  EG, 
which  is  common,  nvA  the  remaining  rectangle  BE.EF  is  equiv- 
alent to  the  square  of  EH.  But  the  rectangle  contained  by 
BE,  EF  is  the  parallelogram  BD,  because  EF  is  equal  to  ED ; 
therefore  BD  is  equivalent  to  the  square  of  EH  ;  but  BD  is 
equivalent  to  the  figure  A;  therefore  the  square  of  EH  is 
<^quivalent  to  A.  The  square  described  on  EH,  therefore,  is  the 
required  square. 

Scho.  Tliis  is  a  particular  case  of  the  sixteenth  proposition 
of  the  fifth  book,  and  an  easy  solution  can  be  effected  by  means 
of  Proportion. 

Pkop.  XIY. — TiiEOR. —  T/ie  siwi  of  the  squares  of  the  sides 
of  a  trapezium  is  equivalent  to  the  sum  of  the  squares  of  the 
diagonals,  tor/ether  icith  four  times  the  square  of  the  straight 
line  joining  the  points  of  bisection  of  the  diagonals. 

Let  ABCD  be  a  trapezium,  having  its  diagonals  AC,  BD 
bisected  in  E  and  F,  aud  let  EF  be  joined  ;  the  squares  of  AB, 
BC,  CD,  DA  are  together  equivalent  to  the 
squares  of  AC,  BD,  together  with  four 
times  the  squai-e  of  EF. 

Join  AF,  EC.  The  squares  of  AB,  AD 
are  together  equivalent  (II.  12,  cor.)  to 
twice  the  sum  of  the  squares  of  DF  and 
AF;  and  the  squares  of  BC,  CD  are  equiv- 
alent to  twice  the  sum  of  the  squares  of  DF  and  CF.  Add 
these  equivalents  together,  and  the  sum  of  the  squares  of  AB, 
BC,  CD,  DA  is  equivalent  to  four  times  the  square  of  DF,  to- 


B 


64  THE    ELEMEXT3    OF  [BOOK    H. 

getlier  with  twice  the  sum  of  the  squares  of  AF  and  CF.  But 
tv/ice  the  squares  of  AF  and  CF  are  equivalent  (11. 12,  cor.)  to  four 
times  the  squares  of  AE  and  EF;  and  (II.  4,  cor.  2)  four  times 
the  square  of  DF  is  equivalent  to  the  square  of  BD,  and  four 
times  the  square  of  AE  to  the  square  of  AC.  Hence  the  squares 
of  AB,  BC,  CD,  DA  are  equivalent  to  the  squares  of  AC  and 
BD,  together  with  four  times  the  square  of  EF.  Therefore,  the 
sum,  etc. 

Cor.  Hence  the  squares  on  the  diagonals  of  a  parallelogram 
are  together  equivalent  to  the  sum  of  the  squares  on  its  sides — 
for  in  the  case  of  a  parallelogram  the  line  EF  vanishes,  as  the 
diagonals  of  a  parallelogram  bisect  each  other. 

/S'cAo.  Hence,  if  we  have  the  sides  and  one  of  the  diagonals 
of  a  parallelogram  in  numbers,  we  can  compute  the  remaining 
diagonal.  Thus,  if  AB,  DC  be  each  =9,  AD,  BC  each  =  7,  and 
AC  =8,  we  have  AB^+BC'+CD=  +  DA=  =  81  +  49+81  +  49  = 
260,  and  AC-r=64.  Taking  the  latter  from  the  fonner,  and  ex- 
tracting the  square  root,  we  find  BD  =  14. 


END    OF   BOOK   SECOND. 


BOOK    THIRD. 

ON  THE  CIRCLE,  AND  LINES  AND  ANGLES  DE- 
PENDING ON  IT,  AND  RECTILINEAL  FIGURES 
DESCRIBED   ABOUT  THE   CIRCLE. 

DEFIXinOIfS. 

1.  A  STRAIGHT  line  is  said  to  touch  a  circle,  or  to  be  a  tangent 
to  it,  when  it  meets  the  circle,  and  being  produced  does  not 
cut  it. 

2.  Circles  are  said  to  touch  one  another,  which  meet,  but  do 
not  cut  one  another. 

3.  In  a  circle,  chords  are  said  to  be  equally  distant  from  the 
center,  when  the  perpendiculars  drawn  to  them  from  the  center 
are  equal. 

4.  And  the  chord  on  which  the  greater  perpendicular  falls, 
is  said  to  \)e  farther  from  the  center. 

5.  An  angle  in  a  segment  of  a  circle  is  an  angle  contained 
by  two  straight  lines  drawn  from  any  point  in  the  arc  of  the 
segment  to  the  extremities  of  its  chord ; 

6.  And  an  angle  is  said  to  stand  upoti  the  arc  intercepted 
between  the  straight  lines  that  contain  the  angle. 

7.  A  sector  of  a  circle  is  a  figure  contained  by  any  arc  of  the 
circle,  and  two  radii  drawn  through  its  extremities. 

8.  A  quadrant  is  a  sector  whose  radii  are  perpendicular  to 
each  other.  It  is  easy  to  show  by  superposition,  that  a  quad- 
rant is  half  of  a  semicircle,  and  therefore  a  fourth  part  of  the 
entire  circle. 

9.  Similar  segments  of  circles  are  those  which  contain  equal 
angles. 

10.  Concentric  circles  are  those  which  have  the  same  center. 

11.  A  regular  polygon  is  equilateral. 

12.  When  the  sides  of  one  rectilineal  figure  pass  through  the 

5 


66  THE   ELEMENTS   OF  [bOOK   HI. 

angular  points  of  another,  the  figures  not  coinciding  with  one 
another,  the  interior  figure  is  said  to  be  inscribed  in  the  exte- 
rior, and  the  exterior  to  be  circutnscribed,  or  described,  about 
the  interior  one. 

13,  When  all  the  angular  points  of  a  rectilineal  figure  are 
upon  the  circumference  of  a  circle,  the  rectilineal  figure  is  said 
to  be  inscribed  in  the  circle,  and  the  circle  to  be  circumscribed, 
or  described,  about  the  rectilineal  figure. 

14.  When  each  side  of  a  rectilineal  figure  touches  a  circle, 
the  rectilineal  figure  is  said  to  be  circumscribed,  or  described, 
about  the  circle,  and  the  circle  to  be  inscribed  in  the  rectilineal 
figure. 

PROPOSITIONS. 

Prop.  I. — Prob. — To  find  the  center  of  a  given  circle,. 

Let   ABC   be   the  given  circle,  and  draw  any  chord  AB. 

Bisect  AB  (I.  6)  by  the  perpendicular 
EDO  drawn  to  the  circumference  on 
both  sides  AB. 

Since  EDC  bisects  AB  (const.)  and  is 

perpendicular  to  AB   (L  6),  the  angles 

AEB   and   ACB   are   also   bisected  by 

EDC,  and  the  subtended  arcs  ACB  and 

AEB   (I.  5)    are    likewise    bisected    by 

EDC;   therefore  the  arcs  CA+AEo 

the   arcs  CB+BE;  hence  CA  +  AE  or 

CE  is  the  semicircumference,  and  the  perpendicular  EDC  is  a 

diameter — consequently  it  passes  through  the  center  of  the 

circle. 

Then  taking  any  other  chord  FIT,  and  in  same  manner  it  can 
be  shown  that  the  perpendicular  MN  which  bisects  the  chord 
is  also  a  diameter,  and  since  all  diameters  pass  through  the 
center  of  the  circle,  the  point  at  which  they  intersect  each 
other,  being  the  only  point  they  have  in  common,  that  point  is 
the  center  of  the  circle. 

Cor.  1.  Hence,  to  find  center  of  any  regular  polygon  (L 
6),  bisect  the  sides  by  perpendiculars  drawn  from  the  points 
of  bisection  (I.  7),  and  the  point  where  the  perpendiculars  inter- 
eect  each  other  is  the  center  of  the  polygon. 


BOOK    m,]  EUCLID   AND   LEGENDRB.  67 

Cor.  2.  In  a  triangle,  straight  lines  drawn  from  the  points 
of  bisection  of  the  three  sides  to  the  opposite  angles  all  pass 
throue'h  the  same  point. 

8cho.  From  the  preceding  coroUarj'^  it  can  be  shown  that 
each  of  the  straight  lines  is  divided  into  two  segments  at  the 
common  point  of  bisection,  of  which  the  segment  nearest  the 
angle  is  double  the  other. 

Prop.  II. — Theor. — If  a  straight  line  drav^nfrom  the  center 
of  a  circle  bisect  a  chord  which  does  not  pass  through  the  ce?i- 
ter,  it  cicts  it  at  right  angles/  and  {2)  if  it  cut  it  at  right  angles, 
it  bisects  it. 

Let  ABC  be  a  circle ;  and  let  ED,  a  straight  line  drawn 
from  the  center  E,  bisect  any  chord  AB,  which  does  not  pass 
through  the  center,  in  the  point  D ;  ED  cuts  AB  at  right 
angles. 

Join  EA,  EB.  Then  in  the  triangles  ADE  BDE,  AD  is 
equal  to  DB,  DE  common,  and  (I.  def  16) 
the  base  EA  is  equal  to  the  base  EB ; 
therefore  (I.  4)  the  angles  ADE,  BDE  are 
equal;  and  consequently  (I.  def  10)  each 
of  them  is  a  right  angle;  wherefore  ED 
cuts  AB  at  right  angles. 

Next,  let  ED  cut  AB  at  right  angles ;  ED  also  bisects  it. 

The  same  construction  being  made,  because  the  radii  EA, 
EB  are  (I.  def  16)  equal,  the  angle  EAD  is  equal  (I.  1,  cor.)  to 
EBD ;  and  the  right  angles  ADE,  BDE  are  equal ;  therefore 
in  the  two  triangles  EAD,  EBD  there  are  two  angles  in  one 
equal  to  two  angles  in  the  other,  each  to  each,  and  the  side 
ED,  which  is  opposite  to  one  of  the  equal  angles  in  each,  is 
common;  therefore  (L  14)  AD  is  equal  to  DB.  If  a  straight 
line,  therefore,  etc.     > 

Cor.  1.  Hence,  in  an  isosceles  triangle,  a  straight  line  drawn 
from  the  vertex  bisecting  the  base  is  perpendicular  to  it ;  and 
a  straight  line  drawn  from  the  vertex  perpendicular  to  the  base, 
bisects  it. 

Cor.  2.  Let  the  straight  line  AB  cut  the  concentric  circles 
ABC,  DEF  in  the  points  A,  D,  E,  B ;  AD  is  equal  to  EB,  and 
AE  to  DB.     From  the  common  center  G,  draw  GH  perpen- 


68 


THE    ELEMEi^TS    OF 


[book  ni. 


dicular  to  AB. 


Then  (III.  2)  AH  is  equal  to  HB,  and  DH  to 
HE.  From  AH  take  DH,  and  from  HB 
take  HE,  and  the  remainders,  AD,  EB, 
are  equal.  To  these  equals  add  DE,  and 
the  sums  AE,  DB  are  equal. 

Cor.  3.  Any  numljer  of  parallel  chords 
in  a  circle  are  all  bisected  by  a  diameter 
perpendicular  to  them. 


Pkop.  hi. — Theok. —  Tico  chords  of  a  circle  which  are  not 
both  diameters,  can  not  bisect  each  other. 

From  the  definition  of  the  circle,  the  center  is  the  only  point 

in  the   circle  which  is  equally  distant 
D  from  all  parts  of  the  circumference  ;  then 

any  chord  which  passes  through  the 
center  is  bisected  at  the  center  ;  the 
diameter  is  the  only  chord  (HI,  1) 
which  passes  through  the  center,  there- 
fore any  other  chord  AB  can  not  bisect 
a  diameter  CD.  And  when  neither  chord 
AB  nor  EF  is  a  diameter,  their  point  of 
intersection  not  being-  the  center  of  the 
circle,  is  unequally  distant  from  the  circumference  ;  therefore 
the  chords  AB  and  EF  are  not  bisected  by  each  other. 


Prop.  IV. — Theor. — Iftico  circles  cut  09ie  another,  they  have 
not  the  same  center. 

Let  ABC  and  DBE  be  two  circles  which  cut  one  another  in 

B;  they  will  not  have  the 
same  center.  For  AC  is  the 
diameter  of  ABC  (III.  1),  and 
DE  is  the  diameter  of  DBE. 
But  the  intersection  of  the 
two  diameters  AC  and  FH 
is  the  center  of  the  circle 
ABC  (in.  1),  and  the  inter- 
section of  the  two  diameters 
LM  and  DE  is  the  center  of  the  circle  DBE.     Now  the  points 


BOOK  m.]  EUCLID  AND  LEGENDRE.  69 

of  intersection  of  these  diameters  are  different,  therefore  ABC 
has  not  the  same  center  with  DBE.  Wherefore,  if  two  cir- 
cles, etc. 

Cor.  1.  Hence,  if  one  circle  touches  another  internally,  they 
have  not  the  same  center. 

Cor.  2.  One  circle  can  not  cut  another  in  more  than  two 
points,  nor  touch  another  in  more  than  one  point. 

Prop.  V. — Theoe. — If  from  any  jjoint  icithin  a  circle, 
which  is  not  the  center,  straight  lines  be  drawn  to  the  circum- 
ference ;  (1)  the  greatest  is  that  tchich  passes  through  the  cen- 
ter, and  (2)  the  continuation  of  that  line  to  the  circumference, 
in  the  opposite  direction,  is  the  least  y  (3)  of  others,  one  nearer 
to  the  line  passing  through  the  center  is  greater  than  one  more 
remote ;  and  (4)  from,  the  same  point  there  can  be  drawn  ordy 
two  equal  straight  lines,  one  upon  each  side  of  either  the  longest 
or  shortest  line,  and  making  equal  angles  vnth  that  line. 

Let  ABCD  be  a  circle,  E  its  center,  and  AD  a  diameter,  in 
which  let  any  point  F  be  taken,  which  is  not  the  center ;  of  all 
the  straight  lines  FA,  FB,  FC,  etc.,  that  can  be  drawn  from  F 
to  the  circumference,  FA  is  the  greatest,  and  FD  the  least ;  and 
of  the  others,  FB  is  greater  than  FC. 

1.  Join  BE,CE.  Then  (I.  21,  cor.)  BE,  EF  are  greater  than 
BF ;  but  AE  is  equal  to  EB ;  therefore  AF,  that  is,  AE,  EF,  is 
greater  than  BF. 

2.  Because  CF,  FE  are  greater  (I.  21,  cor.)  than  EC,  and  EC 
is  equal  to  ED ;  CF,  FE  are  greater  than  ED.     Take  away  the 
common  part  FE,  and  (I.  ax.  5)  the  re- 
mainder CF  is  greater  than  the  remain- 
der FD. 

3.  Again :  because  BE  is  equal  to  CE, 
and  FE  common  to  the  triangles  BEF, 
CEF ;  but  the  angle  BEF  is  greater  than 
CEF ;  therefore  (L  22)  the  base  BF  is 
greater  than  the  base  CF. 

4.  Make  (I.  13)  the  angle  FEH  equal 

to  FEC,  and  join  FH.  Then,  because  CE  is  equal  to  HE,  EF 
common  to  the  two  triangles  CEF,  HEF,  and  the  angle  CEF 
equal  to  the  angle  HEF ;  therefore  (I.  3)  the  base  FC  is  equal 


TO  THE   ELEMENTS   OF  [bOOK    III. 

to  the  base  FH,  and  the  angle  EFC  to  the  angle  EFH. 
But,  besides  FH,  no  other  straight  line  can  be  drawn  from  F 
to  the  circumference  equal  to  FC  (L  ax,  9). 

Prop,  VI, — Theok, — If  from  any  point  without  a  circle 
straight  lines  he  drawn  to  the  circrimference;  (l)  of  those  which 
fall  npon  the  concave  part  of  the  circumference,  the  greatest  is 
that  which  passes  through  the  center ;  unci  (2)  of  the  rest,  one 
nearer  to  the  greatest  is  greater  than  one  more  remote.  (3) 
Hut  of  those  which  fall  upon  the  convex  part,  the  least  is  that 
which  when  produced  passes  through  the  center'  and  (4)  of  the 
rest,  one  nearer  to  the  least  is  less  than  one  more  remote.  And 
(5)  only  two  equal  straight  lines  can  be  drawn  from  the  point 
to  either  part  of  the  circumference,  one  upon  each  side  of  the 
line  passing  through  the  center,  and  making  equal  angles 
with  it. 

Let  ABF  be  a  circle,  M  its  center,  and  D  any  point  without 
it,  from  which  let  the  straight  lines  DA,  DE,  DF  be  drawn  to 
the  circumference.  Of  those  which  fall  upon  the  concave  pait 
of  the  circumference  AEF,  the  greatest  is  DlNLiV,  which  passes 
through  the  center ;  and  a  line  DE  nearer  to  it  is  greater  than 
DF,  one  more  remote.  But  of  those  which  fall  upon  the  con- 
vex circumference  LKG,  the  least  is  DG,  the  external  part  of 
DMA  ;  and  a  line  DK  nearer  to  it  is  less  than  DL,  one  more 
remote, 

1.  Join  ME,  MF,  ML,  MK ;  and  because  MA  is  equal  to  ME, 
add  MD  to  each ;  therefore  AD  is  equal  to  EM,  MD ;  but  (I. 
21,  cor,)  EM,  MD  are  together  greater  than  ED;  therefore,  also, 

AD  is  greater  than  ED. 

2.  Because  ME  is  equal  to  MF,  and 
MD  common  to  the  triangles  EMD, 
FMD,  but  the  angle  EMD  is  greater  than 
FMD  ;  therefore  (L  22)  the  base  ED  is 
greater  than  the  base  FD. 

3,  Because  (I,  21,  cor.)  MK,  KD  are 
greater  than  MD,  and  MK  is  equal  to 
MG,  the  remainder  KG  is  greater  (I,  ax. 
5)  than  the  remainder  GD  ;  that  ie,  GD  L* 
less  than  KD. 


BOOK  m.] 


EUCLID    AND    LEGENDKE. 


71 


4,  Because  MK  is  equal  to  ML,  and  MD  common  to  the 
triangles  KMD,  LMD,  but  the  angle  DMK  less  than  DML; 
therefore  the  base  DK  is  less  (I.  22)  than  the  base  DL. 

5.  Make  (I.  13)  the  angle  DMB  equal  to  DMK,  and  join 
DB.  Then,  because  MK  is  equal  to  MB,  MD  common  to  the 
triangles  KMD,  BMD,  and  the  angle  KMD  equal  to  BMD  ; 
therefore  (I.  3)  the  base  DK  is  equal  to  the  base  DB,  and  the 
angle  MDK  to  the  angle  MDB,  But,  besides  DB,  there  can 
be  no  straight  line  drawn  from  D  to  the  circumference  equal  to 
DK  (I.  ax.  9). 


Prop.  VII. — Theoe. — Equal  chords  in  a  circle  are  equally 
distant  from  the  center  ;  and  (2)  chords  which  are  equally  dis- 
tant from  the  center  are  equal  to  one  another. 

Let  the  chords  AB,  CD,  in  the  circle  ABDC,  be  equal  to  one 
another;  they  are  equally  distant  from  the  center. 

Take  (TIL  1)  E  the  center  of  the  circle,  and  draw  (L  7)  EF, 
EG  perpendiculars  to  AB,  CD  ;  join  also  EA,  EC.  Then,  be- 
cause the  straight  line  EF,  passing  through  the  center,  cuts  the 
chord  AB,  which  does  not  pass  through  the  center,  at  right 
angles,  it  also  (III.  2)  bisects  it ;  where- 
fore AF  is  equal  to  FB,  and  AB  is  dou- 
ble of  AF.  For  the  same  reason,  CD  is 
double  of  CG ;  but  AB  is  equal  to  CD  ; 
therefore  AF  is  equal  (I.  ax.  V)  to  CG. 
Then,  in  the  right-angled  triangles  EF  A, 
EGC,  the  sides  EA,  AF  are  equal  to  the 
sides  EC,  CG,  each  to  each,  therefore 
(L  3)  the  sides  EF,  EG  are  equal.     But 

chords  in  a  circle  are  said  (IIL  def,  3)  to  be  equally  distant 
from  the  center,  when  the  perpendiculars  drawn  to  them  from 
the  center  are  equal;  therefore  AB,  CD  are  equally  distant 
from  the  center. 

Next,  if  the  chords  AB,  CD  be  equally  distant  from  the  cen- 
ter, that  is,  if  FE  be  equal  to  EG,  AB  is  equal  to  CD.  For, 
the  same  construction  being  made,  it  may,  as  before,  be  demon- 
strated that  AB  is  double  of  AF,  and  CD  of  CG  ;  and  because 
tlie  right-angled  triangles  EFA,  EGC  have  the  sides  AE,  EF 
equal  to  CE,  EG,  each  to  each,  the  sides  AF,  CG  are  also  (L  3) 


72  THE   ELEMENTS   OF  [bOOK   m. 

equal  to  one  another.  But  AB  is  double  of  AF,  and  CD  of 
CG ;  wherefore  AB  is  equal  (I.  ax.  6)  to  CD.  Therefore  equal 
chords,  etc. 

Cor.  Hence,  the  diameter  of  a  circle  is  the  greatest  chord ; 
(2)  of  others,  one  nearer  to  the  center  is  greater  than  one  more 
remote  ;  and  (3)  the  greater  is  nearer  to  the  center  than  the 
less. 

Prop.  VIII. — Thkoe. — The  straight  line  drawn  perpendicu- 
lar to  a  diameter  of  a  circle^  through  its  extremity.^  falls  icith- 
out  the  circle  ;  hut  any  other  straight  line  drawn  through  that 
point  cuts  the  circle. 

Let  ABC  be  a  circle,  of  which  D  is  the  center,  and  AB  a 
diameter ;  if  AE  be  drawn  through  A  perpendicular  to  AB,  it 
falls  without  the  circle. 

In  AE  take  any  i:)oint  F ;  and  draw 
DF,  meeting  the  circumference  in  C. 
Because  DAF  is  a  right  angle,  it  is 
greater  (I.  20)  than  DFA;  and  there- 
fore (I.  21)  DF  is  greater  than  DA. 
But  (I.  def.  16)  DA  is  equal  to  DC; 
therefore  DF  is  greater  than  DC,  and 
the  point  F  is  therefore  without  the 
circle ;  and  in  the  same  manner  it  may 
be  shown,  that  any  other  point  in  AF, 
except  the  point  A,  is  without  the  circle. 

Again :  any  other  straight  line  drawn  through  A  cuts  the 
circle. 

Let  AG  be  drawn  in  the  angle  DAF,  and  draw  (I.  8)  DH 
perpendicular  to  AG,  and  meeting  the  circumference  in  C. 
Then,  because  DHA  is  a  right  angle,  and  DAH  less  than  a 
right  angle,  being  a  part  of  DAE,  the  side  DH  is  less  (I.  20) 
than  the  side  DA,  But  (L  def  16)  DK  is  equal  to  DA;  there- 
fore DH  is  less  than  DK ;  the  point  H  is  therefore  within  the 
circle ;  and  AG  cuts  the  circle,  since  its  continuation  through 
A  must  fall  on  the  opposite  side  of  EAL,  and  must  therefore  be 
without  the  circle.     Therefore  the  straight  line,  etc. 

Cor.  From  this  it  is  manifest  that  the  straight  line  which  is 
drawn  at  right  angles  to  a  diameter  of  a  circle  from  its  extrem- 


BOOK  in.] 


EUCLID    AND   LEGENDRE. 


73 


ity  touches  (III.  def.  1)  the  ch'cle;  and  that  it  touches  it  only 
in  one  poiut,  because  at  every  point  except  A,  it  falls  without 
the  circle.  It  is  also  evident,  that  there  can  be  but  one  tan- 
gent at  the  same  point  of  a  circle. 

Pkop.  IX. — Pbob. — From  a  given  pointy  either  -without  a 
given  circle^  or  in  its  circumference^  to  draw  a  straight  line 
touchi^ig  the  circle. 

First :  let  A  be  a  given  point  without  the  given  circle  BCD ; 
it  is  required  to  draw  from  A  a  straight  line  touching  the 
circle. 

Find  (III.  1)  E  the  center  of  the  circle,  and  draw  AE  cutting 
the  circumference  in  D  ;  from  the  center  E,  at  the  distance  EA, 
describe  (I.  post.  3)  the  circle  AFG; 
from  D  draw  (I.  7)  DF  at  right  an- 
gles to  EA;  and  draw  EBF,  and 
join  AB.  AB  touches  the  circle 
BCD. 

Because  E  is  the  center  of  the  cir-  q 
cles,  EA  is  equal  to  EF,  and  ED  to 
EB  ;  therefore  the  two  sides  AE, 
EB  are  equal  to  the  two  FE,  ED, 
each  to  each,  and  they  contain  the 
angle  AEF  common  to  the  two  tri- 
angles AEB,  FED  ;  therefore  the  angle  EBA  is  equal  (I.  3) 
to  EDF,  and  is,  therefore,  a  right  angle,  because  (const.) 
EDF  is  a  right  angle.  Now,  since  EB  is  drawn  from  the  cen- 
ter, it  is  part  of  a  diameter  of  which  B  is  one  extremity ;  but  a 
straight  line  drawn  from  the  extremity  of  a  diameter  at  right 
angles  to  it  touches  (III.  8,  cor.)  the  circle;  therefore  AB 
touches  the  circle ;  and  it  is  drawn  from  the  given  point  A ; 
which  was  to  be  done. 

Secondlj :  if  the  given  point  be  in  the  circumference  of  the 
circle,  as  the  point  D,  draw  DE  to  the  center  E,  and  DF  at 
right  angles  to  DE  ;  DF  touches  (III  8,  cor.)  the  circle. 

Cor,  1.  If  AB  be  produced  to  H,  AH  is  bisected  (III.  2)  in 
B.  Hence  a  chord  in  a  circle  touching  a  concentric  one  is 
bisected  at  the  point  of  contact. 

Scho.  1.  It  is  evident  that  from  any  point  A  without  the  cir- 


74  THE   ELEMENTS   OF  [BOOK   HI. 

cle,  two  tangents  may  be  drawn  to  the  circle,  and  that  these 
are  equal  to  one  another,  being  equal  respectively  to  the  equal 
lines  DF  and  DF'. 

Scho.  2.  The  construction  of  the  first  case  of  this  problem  is 
as  easily  efiected  in  practice,  by  describing  a  circle  on  AE  as 
diameter,  as  its  circumference  will  cut  that  of  the  given  circle 
in  the  points  B  and  B^  The  reason  of  this  will  be  evident 
from  the  twelfth  proposition  of  this  book. 

Cor.  2.  Hence  in  a  given  straight  line  AB,  a  point  may  be 
found  such  that  the  difference  of  its  distances  from  two  given 

points  C,  D,  may   be  equal  to   a   given 
straight  line.     Join  CD,  and  from  D  as 
center,  with   a   radius,   DE  equal  to  the 
given  difference,  describe  a   circle ;  draw 
CP"G  perpendicular  to  AB,  and  make  FG 
D  equal  to  FC  ;  through  C,  G  describe  a  cir- 
cle touching  the  other  circle ;  join  B,  D, 
the  centers  of  the  two  circles,  and  draw  BC ;  BC,  BD  are  evi- 
dently the  required  lines. 

Cor.  3.  In  the  same  manner,  if  a  circle  were  described  fi-om 
D  as  center,  with  the  sum  of  two  given  lines  as  radius,  and  a 
circle  were  described  through  C  and  G  touching  that  circle, 
straight  lines  drawn  from  the  center  of  that  circle  to  C  and  D 
would  be  equal  to  the  given  sum. 

Pkop.  X. — Theor. —  The  angle  at  the  center  of  a  circle  is 
double  of  the  angle  at  the  circuntference.,  xipon  the  same  base^ 
that  is,  upon  the  same  part  of  the  circumference. 

In  the  circle  ABC,  let  BEC  be  an  angle  at  the  center,  and 
BAC  an  angle  at  the  circumference,  which 
have  the  same  arc  BC  for  their  base  ;  BEC 
is  double  of  BAC. 

Draw  AE,  and  produce   it  to  F ;  and 
first,  let    E,   the  center   of  the  circle,  be 
wiiliin  the   angle  BAC.     Because  EA  is 
equal  to  EB,  the  angle  EAB  is  equal  (I.  1, 
"p     ^  cor.)  to  EI>A  ;  therefore  the  angles  EAB, 

EBA  are  together  double  of  EAB  ;  but  (I.  20)  the  angle  BEP 
is  equal  to  EAB,  EBA;  therefore  also  BEF  is  double  of  EAB. 


BOOK    in.]  EUCLID   AND   LKGENDRE.  75 

For  the  same  reason,  the  angle  FEC  is  double  of  the  angle 
EAC ;  therefore  the  whole  angle  BEC  is  double  of  the  whole 
BAG. 

Again :  let  E  the  center  of  the  circle  be  without  the  an^le 

CD  O 

BAG ;  it  may  be  demonstrated,  as  in  the 
first  case,  that  the  angle  FEG  is  double  of 
FAG,  and  that  FEB,  a  part  of  the  first,  is 
double  of  FAB  a  part  of  the  other;  there- 
fore the  remaining  angle  BEC  is  double  of 
the  remaininsj  ansjcle  BAG.  The  angle  at 
the  center,  therefore,  etc. 

Scho.  That  if  two  magnitudes  be  double 
of  two  others,  each  of  each,  the  sum  and  diff'erence  of  the  first 
two  are  respectively  double  of  the  sum  and  difference  of  the 
other  two.  It  is  thus  proved  by  Play  fair:  "Let  A  and  B,  G 
and  D  be  four  magnitudes,  such  that  A  =  2G,  and  B  =  2D; 
then  A+B  =  2(G  +  D).  For  since  A  =  G  +  G,  and  B=D-f-D, 
adding  equals  to  equals,  A4-B  =  (G  +  D)  +  G  +  D  =  2(G  +  D). 
So,  also,  if  A  be  greater  than  B,  and  therefore  G  greater  than 
D,  since  A=:G4-G,  and  B=D4-D,  taking  equals  from  equals, 
A— B  =  (G— D)  +  (G— D),  that  is  A— B  =  2(G— D)." 

The  following  is  an  outline  of  another  proof  of  the  second 
case:  from  the  triangle  BGE  we  have  (I.  20)  BGG=BEG-!-B 
=:BEG4-EAG  (I.  1).  We  have  also  from  the  triangle  AGG, 
in  a  similar  manner,  BGG  =  BAG  +  G  =  BAG  +  EAG  =  2BAG  + 
EAG.  Hence  (I.  ax.  1)  BEG  +  EAG  =  2BAG  +  EAG.  Take 
away  EAG,  etc.  If  in  the  first  diagram,  GE  were  produced  to 
meet  AB,  the  first  case  might  be  proved  in  a  similar  manner. 
The  second  case  might  also  be  proved  by  drawing  from  G  to 
AB  a  straight  line  meeting  it  in  a  point  H,  and  making  Avith 
AC  an  angle  equal  to  BAG.  Then,  by  taking  the  difference 
between  the  equal  angles  EAC,  EGA,  and  the  equal  ones  HAG, 
lie  A,  we  have  EAB  and  EGH  equal,  and  therefore  EBA= 
EGH.  But  EGB=HGG;  and  therefore  (I.  20,  cor.  5)  BEG  = 
BHG=:2BAG.  Tl  e  same  proof,  with  some  obvious  variations, 
would  be  applicable  in  the  first  case. 

There  is  evidently  a  third  case,  viz.,  when  AB  or  AG  passes 
through  the  center;  but  though  this  case  is  not  given  in  a  sep- 
arate form,  its  proof  is  contained  in  that  of  either  of  the  others. 


76 


THE    ELEMENTS    OF 


[book  nic 


PpwOp.  XL — Theor. — In  a  circle,  (1)  the  angle  in  a  semicir- 
cle is  a  right  angle ;  (2)  the  angle  in  a  seginent  greater  than  a 
semicircle  is  acute  ;  and  (3)  the  angle  in  a  segment  less  than  a 
semicircle  is  obtuse. 

Let  ABC  be  a  circle,  of  which  F  is  the  center,  BC  a  diame- 
ter, and  consequently  BAG  a  semicircle ;  and  let  the  segment 
BAD  be  greater,  and  BAE  less  than  a  semicircle ;  the  angle 
BAG  in  the  semicircle  is  a  right  angle ;  but  the  angle  BAD 

in  the  segment  greater  than  a  semicircle 
is  acute  ;  and  the  angle  BAE  in  the  seg- 
ment less  than  a  semicircle  is  obtuse. 

Draw  AF  and  produce  it  to  G.  Then 
(III.  10)  the  angle  BAG  at  the  circum- 
ference, is  half  of  BFG  at  the  center, 
both  standing  on  the  same  arc  BG  ;  and 
for  the  same  reason,  GAG  is  half  of 
GFC.  Therefore  the  whole  angle  BAG  is  half  of  the  angles 
BFG,  GFG ;  and  (I.  9)  these  are  together  equivalent  to  two 
right  angles ;  therefore  DAG  is  a  riglit  angle,  and  it  is  an  angle 
in  a  semicircle.  But  (I.  ax.  9)  the  angle  BAD  is  less,  and  BAE 
greater  than  the  right  angle  BAG  ;  therefore  an  angle  in  a  seg- 
ment greater  than  a  semicircle  is  acute,  and  an  angle  in  a  seg- 
ment less  than  a  semicircle  is  obtuse. 


Pbop.  XII. — Theoe. — If  a  straight  line  touch  a  circle,  the 
straight  line  drawn  from  the  center  to  the  point  of  contact  is 
'perpendicular  to  the  line  touching  the  circle. 
Let  the  straight  line  FH  touch  the  circle  ABGD  at  the  point 

C ;  the  straight  line  GA  drawn  from 
that  point  to  the  center  of  the  circle  is 
perpendicular  to  the  line  touching  the 
circle.  Draw  a  diameter  BD  parallel  to 
FH  (I.  18).  and  with  the  diameter 
BD  as  a  base,  and  the  point  C  as  a  ver- 
tex, make  the  triangle  BGD.  But  BGD 
is  a  right  angle  (III.  11).  FGB  and 
HCD  are  equivalent  to  a  right  angle 
(L  9),  and  because  BD  is  parallel  to  FH  (const.),  FGB  is 
equal  to  GBD,  and  HGD  is  equal  to  CDB  (L  15,  cor.  2) ;  and 


BOOK  in.] 


EUCLID    AND    LEGENDKE. 


77 


at  the  point  C  draw  a  perpendicular  CA  to  FH  (I.  7),  it  will 
also  be  perpendicular  to  BD  (I.  17),  then  FCB  and  BCE  are 
equivalent  to  BCD  (I.  ax.  1)  ;  taking  a^vay  the  common  angle 
BCE,  we  have  FCB  equal  to  ECD ;  but  FCB  is  equal  to 
CBD,  hence  ECD  is  equal  to  CBD  ;  and  in  same  manner  it  can 
be  shown  that  ECB  is  equal  to  CDB.  The  triangles  EBC  and 
EDC  having  EC  common,  the  angle  EBC  equal  to  the  angle 
ECD,  and  the  angles  BEC  and  CED  both  right  angles,  are 
equal  (I.  3),  therefore  EB  is  equal  to  ED,  and  the  diameter  BD 
is  bisected  by  CA,  then  CA  passes  through  the  center  of  the 
circle  (III.  1).     Wherefore,  if  a  straight  line,  etc. 

Cor.  Hence,  conversely,  if  a  straight  line  touch  a  circle,  a 
straight  line  drawn  from  the  point  of  contact,  perpendicular  to 
the  tangent,  passes  through  the  center. 


Prop.  XIII. — Theoe. — If  one  circle  touch  another  inter- 
nally or  externally  in  any  point,  the  straiyht  line  rohich  joins 
their  centers  being  jyroducecl,  passes  through  that  point. 

Let  al)Q  and  DEC  be  two  circles  which  touch  one  another 
internally,  their  centers  will  be  in  the  same  straight  line  with 
their  point  of  contact. 

At  the  point  of  contact  C  draw  the  tangent  FH,  and  at  this 
point  erect  a  perpen- 
dicular CaE  (I.  7).  B  p  D 
The  diameter  of  aJC 
is  perpendicular  to  the 
tangent  at  the  point  of 
contact  (HI.  12),  hence 
Ca  is  the  diameter  of 
a6C,  and  passes 
through  the  center  of 
a5C(IH.  1).     NowFH 

is  also  tangent  to  the  circle  DEC  (const.)  at  the  point  C ; 
hence,  the  perpendicular  CE  (HI.  12)  is  the  diameter  of  DEC, 
and  passes  (HI.  1)  through  the  center  of  DEC.  But  Ca  and 
CE  are  in  the  same  straight  line,  hence  the  centers  of  the  cir- 
cles ahQi  and  DEC,  and  the  point  of  contact  C,  are  in  the 
straight  line  CaE.     Wherefore,  if  one  circle  touch,  etc. 

Or,  let  ABC  and  CDE  be  two  circles  which  touch  one  an- 


78  THE   ELEMENTS   OF  [bOOK   IH. 

Other  externally  in  the  point  C ;  their  centers  will  be  in  the 
same  straight  line  with  the  point  of  contact. 

At  the  point  C  draw  a  tangent  FH,  which  will  be  perpen- 
dicular to  the  diameter  of  ABC  (III.  12).  FH  being  tangent 
at  the  point  of  contact  of  the  circles,  is  also  pei-pendicular  to 
the  diameter  of  CDE  (IIL  12).  ACF  is  a  right  angle  (I.  def. 
10),  and  FCE  for  the  same  reason  is  a  right  angle,  and  both 
equal  to  one  another  (I.  def,  10,  and  ax.  11),  hence  are  two 
right  angles ;  then  (I.  10)  AC  and  CE  form  the  same  straight 
line;  but  AC  passes  through  the  center  of  ABC,  and  CE  passes 
through  the  center  of  CDE,  being  their  respective  diameters 
(IIL  1),  therefore  the  centers  of  the  circles  are  in  the  same 
straight  line  with  the  point  C.  Wherefore,  if  two  circles  touch 
each  other,  etc. 

Prop.  XIV. — Tueor. — Similar  segmevU  of  circles  upon 
equal  bases  are  equal  to  one  another,  and  have  equal  arcs. 

Let  AEB,  CFD  be  similar  segments  of  circles  upon  the  equal 
straight  lines  or  bases,  AB,  CD ;  the  segments  are  equal ;  and 
likewise  the  arcs  AEB,  CFD  are  equal. 

For  if  the  segment  AEB  be  applied  to  the  segment  CFD,  so 

that  the  point  A  may 
E  y_  be     on    C,    and     the 

straight    line   AB    on 


CD,  the  point  B  will 
D  coincide  with  D,  be- 
cause AB  is  equal  to 
CD  ;  therefore  the  straight  line  AB  coinciding  with  CD,  the 
segment  AEB  must  coincide  (I.  def  16)  with  the  segment 
CFD,  and  is  therefore  equal  (I.  ax.  8)  to  it ;  and  the  arcs  AEB, 
CFD  are  equal,  because  they  coincide.  Therefore  similar  seg- 
ments, etc. 

Prop.  XV. — Prob, — A  segment  of  a  circle  being  given,  to 
complete  the  circle  of  lohich  it  is  a  segment. 

Assume  three  points  in  the  arc  of  the  segment,  and  find  (IIL 
1)  the  center  of  the  circle.  From  that  center,  at  the  distance 
between  it  and  any  point  in  the  arc  describe  a  circle,  and  it  will 
evidently  be  the  one  required- 


BOOK  ni.] 


EUCLID  AND  LEGENDEE. 


79 


Prop.  XV L — Treor.  — Ifi  equal  circles,  or  in  the  same  circle, 
equal  a?igles  stand  upon  equal  arcs,  whether  they  are  at  the 
centers  or  the  circumferences. 

Let  ARC,  DEF  be  equal  circles,  having  the  equal  angles 
BGC,  EHF  at  their  centers,  and  BAG,  EDF  at  their  circum- 
ferences ;  the  arc  BKC  is  equal  to  the  arc  ELF. 

Join  BG,  EF ;  and  because  the  circles  ABG,  DEF  are  equal, 
their  radii  are  equal ;  therefore  the  two  sides  BG,  GG  are  equal 


to  tlic  two,  EH,  HF  ;  and  (hyp.)  the  angles  G  and  H  are  equal ; 
therefore  (I.  3)  the  base  BG  is  equal  to  the  base  EF.  Then, 
because  the  angles  A  and  D  are  equal,  the  segment  BAG  is 
similar  (III,  def  8)  to  the  segment  EDF ;  and  they  are  upon 
equal  straight  lines  BG,  EF  ;  but  (III.  14)  similar  segments  of 
circles  upon  equal  straight  lines  have  equal  arcs ;  therefore  the 
arc  BAG  is  equal  to  the  arc  EDF.  But  the  whole  circumfer- 
ence ABG  is  equal  to  the  whole  DEF,  because  the  circles  are 
equal ;  therefore  the  remaining  arc  BKG  is  equal  (I.  ax.  3)  to 
the  remaining  arc  ELF.     Wherefore,  in  equal  circles,  etc. 

Cor.  1.  Gonversely,  in  equal  circles,  or  in  the  same  circle,  the 
angles  which  stand  upon  equal  arcs  are  equal  to  one  another, 
•whether  they  are  at  the  centers  or  the  circumferences.  (I. 
def  19.) 

Cor.  2.  Hence,  in  a  circle,  the  arcs  intercepted  between  par- 
allel chords  are  equal.  For  if  a  straight  line  be  drawn  trans- 
versely, joining  two  extremities  of  the  chords,  it  will  (I.  16) 
make  equal  angles  with  the  chords ;  and  therefore  the  arcs  on 
"which  these  stand  are  eqiial. 

Cor.  3.  Hence,  in  equal  circles,  equal  chords  divide  the  cir- 
cumferences into  parts  which  are  equal,  each  to  each. 


80 


THE   ELEMENTS   OF 


[book  nr. 


D 


Pkop.  XVII. — Peob. — To  bisect  a  given  arc  of  a  circle. 
Let  ADB  be  a  giA-en  arc ;  it  is  required  to  bisect  it. 
Draw  AB,  and  (I.  6  and  7)  bisect  it  in  C,  by  the  perpendicu- 
lar CD ;  the  arc  ABD  is  bisected  in  the  point  D. 

Join  AD,  DB.  Then,  because  AC  is  equal  to  CB,  CD  com- 
mon to  the  triangles  ACD,  BCD,  and  the  angle  ACD  equal  to 

BCD,  each  of  them  being  a  right  angle; 
therefore  (I.  3)  AD  is  equal  to  BD.  But 
(III.  16,  cor.  3),  in  the  same  circle,  equal 
lines  cut  off  equal  arcs,  the  greater  equal 
AC  B       to  the  greater,  and  the  less  to  the  less; 

and  AD,  DB  are  each  of  them  less  than  a 
semicircle,  because  (III.  1)  DC^  or  DC  produced,  passes  througli 
the  center;  wherefore  the  arc  AD  is  equal  to  the  are  DB; 
therefore  the  given  arc  is  bisected  in  D  ;  which  was  to  be  done. 

Prop.  X'N'III. — Theoe. — If  a  straight  line  touch  a  circle, 
and  from  thepoifit  of  contact  a  straight  line  be  drawn  dividing 
the  circle  into  tico  segments  ^  the  angUs  made  by  this  line  with 
the  tangent  are  equivaletit  to  the  angles  which  are  in  the  alter- 
nate segments. 

Let  the  straight  line  DE  touch  the  circle  BAG  in  the  point 
B,  and  let  the  straight  line  BA  be  drawn  dividing  the  circle 
into  the  segments  AGB,  AB  ;  the  angle  ABE  is  equal  to  any  an- 
gle in  the  segment  AGB,  and  the  angle  ABD  to  any  angle  in  AB. 

If  AB  (fig.  1 )  be  perpendicular  to  DE,  it  passes  (III.  1 2,  cor.) 
through  the  center,  and  the  segments  being  therefore  semicir- 
cles, the  angles  in  them  are  (HI.  11)  right  angles,  and  conse- 
quently equal  to  those  which  AB  makes  with  DE. 

But  if  BA  (fig.  2)  be  not  perpendicular  to  DE,  draw  BF  per- 
■     A  F 


D  B  E  D  B  E 

pendicular  to  it ;  join  FA  and  produce  it  to  E;  join  also  CA, 


BOOK  in.] 


EUCLID    AND    LFGKNDRE. 


81 


CB,  C  being  any  point  in  the  arc  ACB.  Then  (TIT.  12,  cor.) 
BF  is  a  diameter,  and  (III.  11,  and  I.  9)  the  angles  BAF,  BAE 
are  riglit  angles.  Therefore,  in  the  triangles  J^AE,  FBE,  the 
angle  E  is  common,  and  the  angles  BAE,  FBE  equal,  being 


right  angles;  wherefore  (I.  20,  cor.  5)  the  remaining  angle 
ABE  is  equal  to  the  remaining  angle  F,  which  is  an  angle  in 
the  remote  or  alternate  segment  BFA. 

Again  :  the  two  angles  ABD,  ABE  are  equal  (I.  9)  to  two 
right  angles ;  and  because  ACBF  is  a  quadrilateral  in  the  circle, 
the  opposite  angles  C  and  F  are  also  equivalent  (I.  20,  cor.  1) 
to  two  right  angles  ;  therefore  (I.  ax.  1)  the  angles  ABD,  ABE 
are  together  equal  to  C  and  F*.  From  these  equals  take  away 
the  angles  ABE  and  F,  which  have  been  proved  to  be  equal ; 
then  (I.  ax.  3)  the  remaining  angle  ABD  is  equal  to  the  re- 
maining angle  C,  which  is  an  angle  iu  the  remote  segment 
ACB.     If,  therefore,  a  straight  line,  etc.. 

ScliO.  1.  The  first  case  is  wanting  in  most  editions  of  Euclid. 
In  the  second  diagram,  FA  and  DB  will  meet  (I.  19)  if  pro- 
duced, the  angles  FBE,  BFA  together  being  evidently  less 
than  two  right  angles. 

Scho.  2.  Let  AB  be  a  fixed  chord,  and  ^ 

through  B  draw  any  other  line  DE  cut- 
ting the  circle  in  C ;  and  join  AC.  Now 
(III.  16,  cor.  1),  wherever  C  is  taken  in 
the  arc  ACB,  the  angle  ACE  is  con- 
stantly of  the  same  magnitude ;  and  so 
also  (I.  9)  is  the  exterior  angle  ACD. 
If  C  be  now  taken  as  coinciding  with  B, 
the  straight  line  DE  will  become  the 
tangent  D'BE',  AC  will  coincide  with  AB,  axid  the  angles 
6 


82  THE   ELEMENTS    OF  [bOOK   HI. 

AGE,  ACD  will  become  ABE',  ABD'.  If  again  C  take  the 
position  C^  the  angles  ACE,  ACD  will  become  AC'^'', 
AC"T>'\  Now,  the  equality  of  ABE',  ACE,  and  of  ABD', 
ACD''  is  what  is  provecl  in  the  eighteenth  proposition,  and 
from  the  equality  of  ACD  and  AC'D"  cor.  2  follows  by  the 
addition  of  ACB. 

Cor.  1.  Angles  in  the  same  segment  of  a  circle  are  equal  to 
one  another. 

Cor.  2.  The  opposite  angles  of  any  quadrilateral  figure  de- 
scribed in  a  circle  are  together  equivalent  to  two  right  angles; 
and  conversely^  if  two  opposite  angles  of  a  quadrilateral  be  to- 
gether equal  to  two  right  angles,  a  circle  may  be  described 
about  it.  * 

Cor.  3.  If  the  circumference  of  a  circle  be  cut  by  two 
straight  lines  which  are  perpendicular  to  one  another,  the 
squares  of  the  four  segments  between  the  point  of  intersection 
of  the  two  lines  and  the  points  in  which  they  meet  the  circum- 
ference, are  together  equal  to  the  square  of  the  diameter. 

Prop.  XIX. — Prob. —  TJpon  a  given  straight  line,  to  describe 
a  segment  of  a  circle  containing  an  angle  equal  to  a  given 
angle. 

Let  AB  be  the  given  straight  line,  and  C  the  given  angle ;  it 
is  required  to  describe  on  AB  a  segment  of  a  circle  containing 
an  angle  equal  to  C. 

First :  if  C  be  a  right  angle,  bisect  (I.  6)  AB  in  F,  and  from 

the  center  F,  at  the  distance  FB,  de- 
scribe the  semicircle  AHB  ;  therefore 
(III.  11)  any  angle  AHB  in  the  semicir- 
cle is  equal  to  the  right  angle  C. 

But  if  C  be  not  a  right  angle,  make 
(I.  13)  the  angle  BAD  equal  to  C,  and 
(I.  7)  from  A  draw  AE  pei-pendicular  to  AD  ;  bisect  (I.  5  and 
6)  AB  by  the  perpendicular  FG,  and  join  GB.  Then,  because 
AF  is  equal  to  FB,  FG  common  to  the  triangles  AFG,  BFG, 
and  the  angle  AFG  equal  to  BFG,  therefore  (I.  3)  AG  is 
equal  to  GB ;  and  the  circle  described  from  the  center  G,  at 
the  distance  GA,  will  pass  through  the  point  B ;  let  this  be  the 
circle  AIIB,     Then,  because  from  the  poiut  A,  the  extremity 


[book   III. 


EUCLID   AND   LEGENDRE. 


83 


of  the  diameter  AE,  AD  is  drawn  at  right  angles  to  AE,  AD 

(III.  8,  cor.)  touches   the  circle;  and   (III.   18)    because  AB, 

drawn  from  the  point  of  contact  A,  cuts 

the  circle,  the  angle  DAB  is  equal  to  any 

angle  in  the  alternate  segment  AHB  ;  but 

DAB  is  equal  to  C ;  therefore  also  C  is 

equal  to  any  angle  in  the  segment  AHB ; 

wherefore  upon   the  given    straight   line 

AB  the  segment  AHB  is  described,  which 

contains  an  angle  equal  to  C  j  which  was 

to  be  done. 

Scho.  It  is  evident  there  may  be  two 
segments  answering  the  conditions  of  the 
problem,  one  on  each  side  of  the  given 
line.     It  is  also  plain,  that  when  C  is  an 
acute  angle,  and  the  segment  is  to  be  above  AB,  G  is  above 
AB;  but  when  obtuse,  it  is  below  it.     It  is  likewise  plain,  that 
the  angle  BAE  is  the  complement  of  the  given  angle  C ;  that 
is,  the  difference  between  it  and  a  right  angle. 

Cor.  Hence   from  a  given  circle  can  be  cut  off  a  segment 
which  shall  contain  an  angle  equal  to  a  given  angle. 


Prop.  XX. — Theor. — If  two  chords  of  a  circle  cut  one  an- 
other, the  rectangle  contained  by  the  segments  of  one  of  them  is 
equal  to  the  rectangle  contained  by  the  segments  of  the  other. 

In  the  circle  LBM,  let  the  two  chords  LM,  BD  cut  one  an- 
other in  the  point  F ;  the  rectangle  ].<F.FM  ^ 
is  equal  to  the  rectangle  BF.FD. 

If  LM,  BD  both  pass  through  the  center, 
80  that  F  is  the  center,  it  is  evident  that 
LF,  FM,  BF,  FD,  being  (I.  def  16)  all 
equal,  the  rectangle  LF.FM  is  equal  to  the 
rectangle  BF.FD. 

But  let  one  of  them,  BD,  pass  through 
the  center  and  cut  the  other,  AC,  which  does  not  pass  through 
the  center,  at  right  angles,  in  the  point  E.     Then,  if  BD  be 
bisected  in  F,  F  is  the  center.     Join  AF;  and  because  BD 
which  passes  through  the  center,  is  perpendicular  to  AC,  AE, 
EC  are  (III.  2)  equal  to  one  another.     Now,  because  BD  is 


84  THE   ELEMENTS   OF  [bOOK   HI. 

divided  equally  in  F,  and  unequally  in  E,  the  rectangle  BE.ED, 
and  the  square  of  EF  are  equivalent  (II.  5)  to  the  square  of  FB  ; 
that  is  (I.  23,  cor.  2),  to  the  square  of  FA.  But  (I.  24,  cor.  I) 
the  squares  of  AE,  EF  are  equivalent  to  the  square  of  FA ; 
therefore  the  rectangle  BE.ED  and  the  square  of  EF  are 
equivalent  to  the  squares  of  AE,  EF.  Take  away  the  common 
square  of  EF,  and  the  remaining  rectangle  BE.ED  is  equivalent 
to  the  remaining  square  of  AE;  that  is,  to  the  rectangle 
AE.EC. 

Next,  let  BD  pass  through  the  center,  and  cut  AC,  which 
does  not  pass  through  the  center,  in  E,  but  not  at  right  angles. 
Then,  as  before,  if  BD  be  bisected  in  F,  F  is  the  center  of  the 

circle.  Join  AF,  and  (I.  8)  draw  FG  per- 
pendicular to  AC  ;  therefore  (III.  2)  AG  is 
equal  to  GC ;  wherefore  (II.  5)  the  rectan- 
gle AE.EC  and  the  square  of  EG  are  equiv- 
alent to  tlie  square  of  AG.  To  each  of 
these  equivaleuts  add  the  square  of  GF; 
therefoi'e  the  rectangle  AE.EC  and  the 
squares  of  EG,  GF  are  equivalent  to  the 
squares  of  AG,  GF ;  but  (I.  24,  coi-.  l)  the  squares  of  EG,  GF  are 
equivalent  to  the  square  of  EF;  and  the  squares  of  AG,  GF  are 
equivalent  to  the  square  of  AF ;  therefore  the  rectangle  AE.EC 
and  the  square  of  EF  are  equivalent  to  the  square  of  AF  ;  that 
is  (I.  28,  cor.  2),  to  the  square  of  FB.  But  (II.  5)  the  square 
of  FB  is  equivalent  to  the  rectangle  BE.ED,  together  with  the 
square  of  EF ;  therefore  the  rectangle  AE.EC  and  the  square 
of  EF  are  equivalent  to  the  rectangle  BE.ED  and  the  square 
of  EF;  take  away  the  common  square  of  EF,  and  the  remain- 
ing rectangle  AE.EC  is  equivalent  to  the  remaining  rectangle 
BE.ED. 

Lastly :  let  neither  of  the  lines  pass  through  the  center,  and 
through  E,  the  point  of  intersection,  draw  a  diameter.  Then, 
the  rectangle  AE.EC  is  equivalent,  as  has  been  shown,  to  the 
rectangle  DE.EB  ;  and,  for  the  same  reason,  the  rectangle  of 
the  other  chord  is  equivalent  to  the  same  rectangle  DPIEB ; 
therefore  (I.  ax.  1)  the  rectangle  AE.EC  is  equivalent  to  the 
rectangle  of  the  other  chord.  If,  therefore,  two  chords  of  a 
circle,  etc. 


BOOK    III.]  EUCLID    AND    LEGENDI  E.  85 

Scho.  The  second  and  third  cases  may  he  thus  demonstrated 
in  one: 

Join  FC.  Then  the  rectangle  AE.EC  is  equivalent  (IT.  5, 
cor.  5)  to  the  difference  of  the  squares  of  AF  and  FE,  or  ot  DF 
and  FE,or  (II.  5,  cor.  l)to  the  rectangle  DE.EB. 

Proportion,  however,  affords  much  the  easiest  method  of 
demonstrating  both  this  proposition  and  the  following. 

Peop.  XXI. — TiiEOR. — If  from  any  point  without  a  c'rcle 
two  straight  li?ies  be  draim,  one  of  which  cuts  the'circle,  and 
the  other  touches  it ;  the  rectangle  contained  hy  the  whole  line 
which  cuts  the  circle^  and  the  />«?•«  of  it  without  the  circle,  is 
equivalent  to  the  square  of  (he  line  which  touches  it. 

Let  D  he  any  point  without  the  circle  ABC,  and  DCA,  DB 
two  straight  lines  drawn  from  it,  of  which  DCA  cuts  the  circle 
in  C  and  A,  and  DB  touches  it  in  B ;  the  rectangle  AD.DC  is 
equivalent  to  the  square  of  DB. 

There  are  two  cases — first :  let  DCA  pass  through  the  center 
E,  and  join  EB.  Therefore  (III.  12)  the 
angle  EBD  is  a  right  angle;  and  (II.  6) 
"because  the  straight  line  AC  is  bisected  in 
E,  and  produced  to  D,  the  rectangle  AD.DC 
and  the  square  of  EC  are  together  equiva- 
lent to  the  square  of  ED  ;  and  CE  is  equal 
to  EB ;  therefore  the  rectangle  AD.DC  and 
the  square  of  EB  are  equivalent  to  the 
square  of  ED.  But  (I.  24,  cor.  l)  the 
square  of  ED  is  equivalent  to  the  squares  a 

of  EB,  BD,  because  EBD  is  a  right  angle ; 
therefore  the   rectangle   AD.DC   and   the  square  of  EB  are 
equivalent  to  the  squares  of  EB,  BD  ;  take  away  the  common 
square  of  EB,  and  the  remaining  rectangle  AD.DC  is  equiva- 
lent to  the  square  of  the  tangent  DB. 

Second :  if  DCA  do  not  pass  through  the  center,  take  (III.  1) 
the  center  E,  and  draw  (I.  8)  EF  perpendicular  to  AC,  and 
join  EB,  EC,  ED.  Then,  because  the  straight  line  EF,  which 
passes  through  the  center,  is  perpendicular  to  the  chord  AC, 
AF  is  equal  (III.  2)  to  FC.  And  (II.  6)  because  AC  is  bisected 
iu  F,  and  produced  to  D,  the  rectangle  AD.DC  and  the  square 


86 


THE   ELEMENTS   OF 


[book  ni. 

of  FC  are  equivalent  to  the  square  of  FD.     To  each  of  these 

equals  add  the  square  of  FE  ;  therefore  the  rectangle  AD.DC, 

and  the  squares  of  CF,  FE  are  equivalent  to 

the  squares  of  DF,  FE  ;  but  (I.  24,  cor.  1) 

the  square  of  ED  is  equivalent  to  the  squares 

of  DF,  FE,  because  EFD  is  a  right  angle ; 

and  the  square  of  EC,  or  (I,  23,  cor.  2)  of 

EB,  is  equivalent  to  the  squares  of  CF,  FE ; 

therefore  the   rectangle    AD.DC    and   the 

square  of  EB  are  equivalent  to  the  square 

of  ED.     But  (I.  24,  cor.  1)  the  squares  of 

EB,  BD  are  equivalent  to  the  square  of  ED 

because  EBD  is  a  right  angle;  therefore  the 

rectangle   AD.DC   and  the  square  of  EB 

are  equivalent   to  the  squares   of  EB.BD.      Take  away  the 

common    square   of  EB  ;    therefore  the   remaining   rectangle 

AD.DC  is  equivalent  to  the  square  of  DB ;  wherefore,  if  from 

any  point,  etc. 

Scho.  The  second  case  may  be  demonstrated  more  briefly 
thus:  Join  AE.  Then  the  rectangle  AD.DC  is  equal  (II.  5, 
cor.  5)  to  the  difference  of  the  squares  of  ED  and  EC,  or  of  ED 
and  EB,  or  (I.  24,  cor.  1)  to  the  square  of  DB. 

Cor.  1.  If  from  any  point  without  a  circle  there  be  drawn 
two  straight  lines  cutting  it,  as  AB,  AC,  the 
rectangles  contained  by  the  whole  lines  and 
the  parts  of  them  without  the  circle  are 
equivalent  to  one  another,  viz.,  the  rectangle 
BA.AE  to  the  rectangle  CA.AF;  for  each 
rectangle  is  equivalent  to  the  square  of  the 
tangent  AD. 

Cor.  2.  If  two  straight  lines  intersect  each 
other,  so  that  the  rectangle  under  the  seg- 
ments of  one  of  them  is  equal  to  the  rectangle 
under   the  segments  of  the  other,  their  ex- 
tremities lie  in  the  circumference  of  a  circle. 

Scho.  By  means  of  the  first  corollary,  it  would  be  shown  in 
a  similar  manner,  that  if  two  straight  lines  meet  in  a  point,  and 
if  they  be  so  divided  that  the  rectangle  under  one  of  them  and 
its  segment  next  the  common  point  is  equal  to  the  rectangle 


BOOK.  III.] 


EUCLID    AND   LEGENDRE. 


87 


under  the  otlier  and  its  corresponding  segment,  the  points  of 
section  and  the  extremities  remote  from  the  common  point  lie 
in  the  circumference  of  the  same  circle. 


D 


Pkop.  XXII. — TiiEOR. — If  frotn  a  point  without  a  circle 
there  be  draw:i  two  straight  lines,  one  of  ichich  cuts  the  ctrcle^ 
and  the  other  meets  it ;  and  if  the  rectangle  contained  by  the 
whole  line  which  cuts  the  circle,  and  the  part  of  it  without  the 
circle,  be  equivalent  to  the  square  of  the  line  which  meets  it,  the 
line  which  meets  the  circle  touches  it. 

If  from  a  point  without  the  circle  ABC  two  straight  lines 
DCA  and  DB  be  drawn,  of  which  DCA  cuts  the  circle,  and  DB 
meets  it ;  and  if  tlie  rectangle  AD.DC  be  equivalent  to  the 
square  of  DB  ;  DB  touches  the  circle. 

Draw  (III.  9)  the  straight  line  DE  touching  the  circle  ABC ; 
find  (III.  1)  the  center  F;  and  join  FE,  FB, 
FD.  Then  (III.  12)  FED  is  a  right  angle; 
and  (fll.  21)  because  DE  touches  the  circle 
ABC,  and  DCA  cuts  it,  the  rectangle  AD.DC 
is  equivalent  to  the  square  of  DE.  But  (hyp.) 
the  rectangle  AD.DC  is  equivalent  to  the 
square  of  DB  ;  therefore  the  square  of  DE  is 
equivalent  to  the  square  of  DB ;  and  the 
straight  line  DE  equal  (I.  23,  cor.  3)  to  the 
straight  line  DB.  But  FE  is  equal  to  FB, 
and  the  base  FD  is  common  to  the  two  tri- 
angles DEF,  DBF ;  therefore  (I.  4)  the  angle  DEF  is  equal  to 
DBF;  but  DEF  is  a  right  angle;  therefore,  also,  DBF  is  a 
right  angle  ;  and  FB  is  a  part  of  a  diameter,  and  the  straight 
line  which  is  drawn  at  riiiht  angles  to  a  diameter,  from  its  ex- 
tremity,  touches  (III.  8,  cor.)  the  circle ;  therefore  DB  touches 
the  circle  ABC.     Wherefore,  if  from  a  point,  etc. 

Prop.  XXIII. — Prob, —  To  divide  a  given  straight  line  into 
two  j)arts,  such  that  the  sjuare  of  one  of  them  may  be  equivalent 
to  the  rectangle  contained  by  the  other,  and  a  given  straight  line. 

Let  AB,  AC  be  two  given  straight  lines  ;  it  is  required  to 
divide  AB  into  two  parts,  such  that  the  square  of  one  of  them 
may  be  equivalent  to  the  rectangle  under  AC  and  the  other. 


88 


THE   ELEMENTS    OF 


[book  m. 


On  CB,  the  sum  of  the  given  lines,  describe  the  semicircle 
CDB,  and  draw  AD  jterpendicular  to  CB  ;  bisect  CA  in  E,  and 
join  DE ;  and  make  EF  equal  to  ED ;  then  the  square  of  AF 
is  equivalent  to  the  rectangle  AC.FB. 

Describe  the  semicircle  CGA,  cutting  ED  in  G,  and  join  GF. 
Then,  because  FE,  EG  are  respectively 
equal  to  DE,  EA,  and  the  angle  FED 
common,  GF  is  equal  to  AD,  and  the  an- 
gle EGF  to  EAD,  whicli  is  a  right  angle, 
and  therefore  GF  touches  the  circle  CGA. 
Hence  (III.  20)  the  rectangle  CA.AB  is 
equivalent  to  the  square  of  AD,  or  of  FG, 

or  (HI.  21)  to  the  rectangle  CF.FA.  But  (11.  1)  the  rectanglo 
CA.AB  is  equivalent  to  the  two  CA.AF,  CA.FB,  and  (II.  3) 
the  rectangle  CF.FA  is  equivalent  to  CA.AF  and  the  square  of 
AF.  From  these  equivalents  take  away  the  rectangle  CA.AF, 
and  there  remains  the  rectangle  CA.FB  equivalent  to  the 
square  of  AF.  • 

Scho.  The  tenth  proposition  of  the  second  book  is  the  par- 
ticular case  of  this  problem  in  which  the  given  lines  are  equal. 


Prop.  XXIV. — Prob. — To  draw  a  common  tangent  to  two 
given  circles. 

Let  BDC,  FUG  be  given  circles ;  it  is  required  to  di-aw  a 
common  tangent  to  them. 

Join  their  centers  A,  E,  and  make  BK  equal  to  FE ;  from  A 


as  center,  with  AK  as  radius,  describe  a  circle  cutting  another, 
described  on  AE  as  diameter,  in  M ;  draw  AM,  meeting  tlie 
circle  BCD  in  D  ;  and  draw  Eli  parallel  to  AM;  join  Dil;  it 
touches  both  the  given  circles. 


BOOK    III.] 


EUCLID    AND   LEGENDRE. 


89 


For  MD,  EII,  which  (const.)  are  parallel,  are  equal  to  one 
another,  because  each  of  them  is  equal  to  KB;  therefore  (I.  15, 
cor.  B)  Dll  is  parallel  to  ]\1E.  Now  (III.  11),  the  angle  AME 
in  a  semicircle  is  a  right  angle;  and  therefore  (I.  16)  the  angles 
ADII,  DUE  are  right  angles,  and  (III.  8,  cor.)  DH  touches 
both  the  ciicles,  since  it  is  perpendicular  to  the  radii  AD,  EH. 

ScJio.  In  the  figure  the  circles  lie  on  the  same  side  of  the 
tangent,  which  is  therefore  exterior  to  them  ;  but  the  tangent 
can  be  transverse,  or  lie  between  the  circles.  It  is  plain  also, 
tliat  in  these  figures,  by  using  the  point  L  instead  of  M,  an- 
other exterior  and  another  transverse  tangent  would  be  ob- 
tained ;  and  this  will  always  be  so,  when  each  of  the  circles 
lies  wholly  without  the  other,  and  does  not  touch  it.  But  if 
the  circles  touch  one  another  externally,  the  two  transverse 
tangents  coalesce  into  a  single  line  passing  through  their  point 
of  contact ;  if  they  cut  one  another,  there  will  be  two  exterior 
tangents,  but  no  transverse  one ;  if  one  of  the  circles  touch  the 
other  internally,  they  carf  have  only  one  common  tangent,  and 
this  passes  through  their  point  of  contact ;  and,  lastly,  if  one  of 
them  lie  wholly  within  the  other  without  touching  it,  they  can 
have  no  common  tangent. 

If  the  circles  be  equal,  the  points  of  contact  of  the  exterior 
common  tangents  are  the  extremities  of  the  diameters  perpen- 
dicular to  the  line  joining  the  centers  ;  for  (I.  15,  cor.  3,  and  1 6) 
the  lines  connecting  the  extremities  of  these  toward  the  same 
parts  are  perpendicular  to  the  diameters,  and  therefore  (III.  8, 
cor.)  they  touch  the  circles. 

Prop.  XXV.— Prob.— 7b  insa-ibe  a7iy  regular  polygon  in  a 
given  circle. 

Let  ABDC  be  the  given  circle  ;  it  is 
required  to  inscribe  any  polyg<m  in  it. 

Since  (I.  9,  cor.)  all  the  angles  formed 
by  any  number  of  straight  lines  inter- 
secting each  other  in  a  common  point 
are  equivalent  to  four  right  angles,  and 
(I,  20,  cor.  1)  any  rectilinear  figure  can 
"be  divided  into  as  many  triangles  as  the 
figure  has  sides,  by  straight  lines  joining  the  extremities  of  the 


90  THE   ELEMENTS   OF  [BOOK   m. 

sides  with  a  poiut  within  the  figure — a  regular  polygon  being 
equilateral — the  straight  lines  connecting  the  extremities  of 
those  sides  with  the  center  of  the  regular  polygon  will  divide 
the  regular  polygon  into  equal  triangles  ;  for  if  ABC  be  a  regu- 
lar polygon  of  three  sides,  AB  is  equal  BC,  and  also  equal  to 
CA,  and  draw  from  E  (III.  1,  cor.  1),  the  center  of  the  polygon, 
EA,  EB,  and  EC.  Now  the  triangles  AEB,  BEC,  and  CEA, 
having  their  bases  equal  (hyp.),  their  other  sides  common, 
(const.)  are  equal;  therefore  (I.  4)  the  angles  AEB,  BEC,  and 
CEA  are  equal ;  but  those  angles  are  equivalent  to  form  right 
angles  (I.  20,  cor.  1) ;  hence,  each  is  one  third  of  four  right 
angles,  or  two  thirds  of  two  right  angles ;  consequently,  a  reg- 
ular polygon  can  be  inscribed  in  a  circle  by  making  an  angle 
on  the  diameter  (I.  13)  with  the  center  of  the  circle  the  vertex 
of  the  angle,  equal  to  that  part  of  four  right  angles  that  the 
regular  polygon  has  sides,  viz. :  if  the  regular  polygon  has  four 
sides,  each  angle  at  the  center  will  be  one  fourth  of  four  right 
angles ;  if  five  sides,  one  fifth  of  fourS-ight  angles;  and  so  on. 
Or,  by  bisecting  (I.  5)  the  angle  or  the  arc  (III.  11),  a  regular 
polygon  of  double  the  number  of  sides  can  be  inscribed  in  the 
circle.  Having  by  these  means  the  angle  at  the  center  of  a 
regular  polygon,  the  chord  of  the  arc  intercepted  by  the  sides 
of  the  angles  will  be  the  side  of  the  regular  polygon  required, 
from  which  (I.  12  and  13)  the  regular  polygon  can  be  inscribed 
in  the  given  circle,  which  was  to  be  done. 

Cor.  1.  If  the  sides  of  the  angles  be  produced  beyond  the 
circumference  (I.  post.  2),  and  parallels  to  the  exterior  sides  of 
the  polygon  (I.  18)  be  drawn  touching  the  circle  (III.  9),  then 
a  similar  and  regular  polygon  can  be  described  about  the  given 
circle. 

Cor,  2.  Find  the  center  of  a  given  regular  polygon  (III.  1, 
cor.  1),  and  a  circle  can  be  inscribed  in  the  polygon,  or  circum- 
scribed about  it,  by  taking  the  straight  line  drawn  from  the 
center  of  the  polygon  to  an  extremity  of  the  side  of  the  poly- 
gon i'or  a  radius  lor  the  circumscribed  circle,  and  a  line  drawn 
from  the  center  to  the  point  of  bisection  of  the  side  for  a  radius 
of  the  inscribed  circle. 

Cor.  3.  Since  the  triangles  AEB,  BEC,  and  AEC  are  equiv- 
alent respectively  to  the  rectangles  (I.  23,  cor.  6)  under  the  ra- 


BOOK   in.]  EUCLID   AND   LEGENDRE.  01 

dius  of  an  inscribed  circle  and  the  halves  of  AB,  BC,  and  CA, 
it  follows  (IL  1)  that  the  area  of  ABC  is  equivalent  to  tlie  rect- 
angle under  the  radius  and  half  the  perimeter.  Hence,  if  the 
sides  be  given  in  numbers,  the  length  of  the  radius  may  be 
computed  by  calculating  (II.  12,  scho.)  the  area,  and  dividing 
it  by  half  the  sum  of  the  sides,  or  its  double  by  the  sum  of  the 
sides. 

Cor.  4.  Since  the  angles  formed  about  the  center  of  a  circle 
are  together  equivalent  to  four  right  angles  (I.  9,  cor.),  and 
since  the  angles  of  a  triangle  are  together  equivalent  to  two 
right  angles  (I.  20),  it  follows  that  when  an  equiangular  trian- 
gle is  formed  (I.  2)  having  a  vertex  at  the  center  and  the  radius 
for  a  side,  each  angle  of  the  triangle  is  one  tliird  of  two  riglit 
angles,  or  one  sixth  of  four  right  angles ;  hence,  six  equal 
angles  can  be  formed  (I.  1.3)  about  the  center  of  a  circle,  and 
since  the  sides  about  tliese  angles  are  intercepted  by  the  cir- 
cumference (const.),  they  are  equal  (I.  def  16)  ;  and  (I.  2,  cor.) 
an  equiangular  triangle  being  also  equilateral,  the  side  of  the 
triangle  opposite  the  angle  at  the  center  of  the  circle  is  equal 
to  the  radius  of  the  circle  (T.  ax.  1)  ;  therefore  the  radius  can 
be  made  to  subtend  six  equal  arcs  of  the  circumference. 

Cor.  5.  When  in  the  case  of  a  general  proposition  to  descrV^e 
a  circle  touching  three  given  straight  Imes  ichich  do  vot  pass 
through  the  same  point,  and  which  are  not  all  parallel  to  one 
another.  If  two  of  the  lines  be  parallel,  there  may  evideiidy 
be  two  equal  circles,  one  on  each  side  of  the  line  falling  on  the 
parallels,  each  of  which  will  touch  the  three  given  lines,  and 
their  centers  will  be  the  intersections  of  the  lines  bisectimr  the 
angles  made  by  the  parallels  with  the  third  line.  But  if  the 
lines  form  a  triangle  by  their  intersections,  there  will  be  four 
circles  touching  them;  one,  inscribed,  and  the  others  each 
touching  one  side  externally  and  the  other  two  produced.  The 
centers  of  the  external  circles  will  be  the  intersections  of  the 
lines  bisecting  two  exterior  angles;  and  the  line  bisecting  the 
remote  interior  angle  will  pass  through  the  same  point.  The 
method  of  proof  is  tl.e  same  as  that  given  in  the  proposition. 

Cor.  6.  If  straight  lines  be  drawn  from  the  center  of  one  of 
the  external  circles  to  the  vertices  of  the  triangle,  the  three  tri- 
angles formed  by  the  sides  of  the  triangle  and  the  straight  lines 


92  THE   ELEMENTS   OF  [bOOK   HI. 

are  respectively  equivalent  to  the  rectangles  (I.  15,  cor.  4) 
under  the  radius  of  that  circle  and  halves  of  tlie  sides  of  the 
triangle.  And  if  the  triangle  formed  by  the  b.ide  of  the  original 
triangle  nearest  the  center  of  the  circle,  and  the  lines  drawn  to 
the  vertices  at  the  extremities  of  that  side,  be  taken  from  the 
sum  of  the  two  other  triangles,  there  remains  the  original  tri- 
angle equivalent  to  the  rectangle  under  the  radius,  and  the  ex- 
cess of  half  the  sum  of  the  two  other  sides  of  the  orin-inal  trian- 
gle  above  half  the  side  nearest  the  center  of  the  circle,  or,  which 
is  the  same  thing,  to  the  rectangle  under  the  radius,  and  the 
excess  of  half  the  perimeter  of  the  original  triangle  above  the 
Bide  nearest  the  center  of  the  circle.  The  radius,  therefore,  of 
any  of  the  external  circles  may  be  computed  by  dividing  the 
area  of  the  original  triangle  by  the  excess  of  half  its  perimeter 
above  the  side  which  the  circle  touches  externally. 

Scho.  The  polygons  considered  in  this  proposition,  and  those 
which  may  be  derived  from  them  by  the  process  of  bisecting 
the  angles  or  arcs  subtending  the  sides  of  the  polygons,  are  the 
only  ones  till  lately  which  geometers  have  been  able  to  de- 
scribe by  elementary  geometry,  that  is,  by  means  of  the 
straight  line  and  circle.  M.  Gauss,  of  Gottingen,  in  \n?,  Disqui- 
sitioiies  Arithmeticoe,  has  shown  that  by  elementary  geometry, 
every  regular  poly. on  may  be  inscribed  in  a  circle, the  number 
of  whose  sides  is  a  power  of  2  increased  by  unity,  and  is  a 
prime  number,  or  a  number  which  can  not  be  produced  by  the 
multiplication  of  two  whole  numbers,  such  as  17 — the  fourth 
power  of  2  increased  by  unity — and  polygons  of  257  and  C5537 
sides.  But  the  investigation  is  too  complex  and  difficult  for  an 
ordinary  school  text-book. 


END    OP   b60K   third. 


BOOK  FOURTH.* 

THE  GENERAL  THEORY  OF  PROPORTION". 

DEFINITIONS. 

1.  A  LESS  number  or  magnitude  is  said  to  measure  a  greater, 
or  to  be  a  measure^  a  part^  or  a  suhmuUxple  of  tlie  greater,  when 
the  less  is  contained  a  certain  number  of  times,  exactly,  in  the 
greater;  and 

2.  The  greater  is  said  to  be  a  multiple  of  the  less. 

3.  Magnitudes  which  can  be  compared  in  respect  of  quantity, 
that  is,  which  are  either  equal  to  one  another,  or  unequal,  are 
Baid  to  be  of  the  sam,e  kind,  or  homogeneous. 

Scho.  1.  Thus,  lines,  whether  straight  or  curved,  are  magni- 
tudes of  the  same  kind,  or  are  homogeneous,  since  they  may  be 
equal  or  unequal.  In  like  manner,  surfaces,  solids,  and  angles 
form  three  other  classes  of  homogeneous  magnitudes.  On  the 
contrary,  lines  and  surfaces,  lines  and  angles,  surfaces  and 
solids,  etc.,  are  heterogeneous.  Thus,  it  is  obviously  improper 
to  say,  that  the  side  and  area  of  a  square  are  equal  to  one  an- 
other, or  are  unequal.  So  likewise  we  cannot  say  that  the  area 
and  one  of  the  angles  of  a  triangle  are  equal  to  each  other,  or 
are  unequal ;  and  they  are  therefore  heterogeneous. 

4.  If  there  be  two  magnitudes  of  the  same  kind,  the  relation 
which  one  of  them  bears  to  the  other  in  respect  of  quantity,  is 
called  its  ratio  to  the  other. 

The  first  term,  or  magnitude,  is  called  the  antecedent  of  the 
ratio,  and  the  second  the  consequent. 

5.  If  there  be  four  magnitudes,  and  if  any  like  multiples 
whatever  be  taken  of  the  first  and  third,  and  any  whatever  of 

*  The  accurate  but  prolix  method  of  Euclid  is  substituted  by  the  fol- 
lowing more  concise  method,  by  employing  the  notations  and  simple 
principles  of  Algebra.    See  Sup.  to  Book  V.  Thombon's  Euclid. 


^^  THE   ELEMENTS   OF  [bOOK   IV. 

the  soconrl  anfl  fourth  ;  and  if,  according  as  the  multiple  of  the 
first  is  gi-eater  than  the  multiple  of  the  second,  equal  to  it,  or 
less,  the  multiple  of  the  third  is  also  greater  than  the  multiple 
of  the  fourth,  equal  to  it,  or  less ;  then  the  first  of  the  magni- 
tudes is  said  to  have  to  the  second  the  same  ratio  that  the  third 
has  to  the  fourth. 

6.  Magnitudes  which  have  the  same  ratio  are  called  propor- 
tionals;  and  equality  or  identity  of  ratios  constitutes /irojoor- 
tion  or  analogy. 

When  magnitudes  are  proportionals,  the  relation  is  expressed 
briefly  by  saying,  that  the  first  is  to  the  second,  a«  the  third  to 
the  fourth,  the  fifth  to  the  sixth,  and  so  on. 

1.  When  of  the  multiples  of  four  magnitudes,  taken  as  in  the 
fifth  definition,  the  multiple  of  the  first  is  greater  than  that  of 
the  second,  but  the  multiple  of  the  third  is  not  greater  than 
that  of  the  fourth  ;  then  the  first  is  said  to  have  to  the  second 
a  greater  ratio  than  the  third  has  to  the  fourth  ;  and,  on  the 
conti-ary,  the  third  is  said  to  have  to  the  iomth  a.  less  ratio  than 
the  first  has  to  the  second. 

8.  When  there  are  three  or  more  magnitudes  of  the  same 
kind,  such  that  the  ratios  of  the  first  to  the  second,  of  the  sec- 
ond to  the  third,  and  so  on,  whatever  may  be  their  number,  are 
all  equal;  the  magnitudes  are  said  to  be  continual  propor- 
tionals. 

9.  The  second  of  three  continual  proportionals  is  said  to  be  a 
mean  proportional  between  the  other  two. 

10.  When  there  is  any  number  of  magnitudes  of  the  same 
kind,  greater  than  two,  the  first  is  said  to  have  to  the  last  the 
ratio  compounded  of  the  ratio  which  the  first  has  to  the  second, 
and  of  the  latio  which  the  second  has  to  the  third,  and  of  the 
ratio  which  the  third  has  to  the  fourth,  and  so  on  to  the  last 
magnitude. 

For  exam].]e,  if  A,  B,  C,  D  be  four  magnitudes  of  the  same 
kind,  the  first,  .>,  is  said  to  have  to  the  last,  D,  the  ratio  com- 
pounded of  tlie  ratios  of  A  to  B,  B  to  C,  and  C  to  D. 

11.  When  iln-ee  magnitudes  are  continual  proportionals,  the 
ratio  of  tlie  first  to  the  third  is  said  to  be  duplicate  of  the  ratio 
of  the  first  Xo  the  second,  or  of  the  second  to  the  third. 

12.  When  iour  magnitudes  are  continual  proportionals,  the 


BOOK  rv.]         EUCLID  AND  LEGENDEE.  95 

ratio  of  the  first  to  the  fourth  is  said  to  be  triplicate  of  the  ratio 
of  the  first  to  the  second,  of  the  ratio  of  the  second  to  the  third, 
or  of  tlie  ratio  of  the  third  to  the  fourth. 

Scho.  2.  In  continual  proportionals,  by  their  own  nature,  and 
that  of  compound  ratio,  the  ratio  of  the  first  to  the  third  is 
compounded  of  two  equal  ratios ;  and  the  ratio  of  the  first  to 
the  fourth,  of  three  equal  ratios ;  and  hence  we  see  the  reason 
and  the  propriety  of  calling  the  first  duplicate  ratio,  and  the 
second  triplicate.  It  is  plain,  that  on  similar  principles,  the 
ratio  of  the  first  to  the  fifth  would  be  said  to  be  quadruplicate 
of  the  ratio  of  the  first  to  the  second,  of  the  second  to  the  third, 
etc.,  and  thus  we  might  form  other  similar  terms  at  pleasure. 

The  terras  subduplicate,  subtriplicate^  and  sesquiplicate^ 
which  are  sometimes  employed  by  mathematical  wiiters,  are 
easily  understood  after  the  explanations  given  above.  In  con- 
tinual proportionals,  the  ratio  of  the  first  terra  to  the  second  is 
said  to  be  subcJuplicate  of  the  ratio  of  the  first  to  the  third,  and 
subtriplicate  of  that  of  the  first  to  the  fourth.  Again  :  if  there 
be  four  continual  proportionals,  the  ratio  of  the  first  to  the 
fourth  is  said  to  be  sesquiplicate  of  the  ratio  of  the  first  to  the 
third ;  or,  which  amounts  to  the  same,  the  ratio  which  is  com- 
pounded of  another  ratio  and  its  subduplicate,  is  sesquiplicate 
of  that  ratio. 

1 3.  In  proportionals,  the  antecedent  terms  are  called  homolo- 
gous to  one  another,  as  also  the  consequents  to  one  another. 

Geometers  make  iise  of  the  following  technical  words  to 
denote  diiferent  modes  of  deriving  one  proportion  from  an- 
other, by  changing  either  the  order  or  the  magnitudes  of  the 
terms. 

14.  Alternately :  this  word  is  used  when  there  are  four  pro- 
portionals of  the  same  kind  ;  and  it  is  inferred  that  the  first  has 
the  same  ratio  to  the  third  which  the  second  has  to  the  fourth ; 
or  that  the  first  is  to  the  third  as  the  second  to  the  fourth ;  as 
is  shown  in  the  fourth  proposition  of  this  book. 

15.  By  inversion:  when  there  are  four  proportionals,  and  it 
is  inferred  that  the  second  is  to  the  first  as  the  fourth  to  the 
third.     Prop.  3,  Book  IV. 

16.  By  composition:  when  there  are  four  proportionals,  and 
it  is  inferred,  that  the  first,  together  with  the  second,  is  to  the 


96  THE   ELEMENTS   OF  [BOOK   IV. 

second  as  the  tliii-fl,  together  with  the  fourth  is  to  the  fourth. 
Tenth  Prop.,  Book  IV. 

17.  By  division  :  when  there  are  four  proportionals,  avd  it 
is  inferred  that  the  exjcess  of  the  first  above  the  second  is  to  the 
second,  as  the  excess  of  the  third  above  the  fourth  is  to  the 
foui-th.     Tenth  Prop.,  Book  IV. 

18.  My  coyiveraion  :  when  there  are  four  proportionals,  and 
it  is  inferred  that  the  first  is  to  its  excess  above  tlie  second,  as 
the  third  to  its  excess  above  the  fourth.  Eleventh  Prop.,  Book 
IV. 

Sc'ho.  The  substance  of  the  five  preceding  definitions  may  be 
exhibited  briefly  in  the  following  manner,  the  signs  +  and  — 
denoting  addition  and  subtraction,  as  has  been  explained 
already  at  the  beginning  of  the  second  book: 

Let  A:  B::  C:D; 
Alternately,  A  :  C  : :  B  :  D  ; 

By  inversion,  B  :  A  : :  D  :  C  ^ 

By  composition,  A  +  B  :  B  : :  C  +  D  :  D  ; 

By  division.  A— B  :  B  : :  C— D  :  D  ; 

By  conversion,  A  :  A — B  : :  C  :  C — D. 

19.  Ex  mquo^  or  ex  equali  (scil.  distantid)^  from  equality  of 
distance:  when  there  is  any  number  of  magnitudes  more  than 
two,  and  as  many  others,  which,  taken  two  in  the  one  rank,  and 
two  in  the  other,  in  direct  order,  have  the  same  ratio;  and  it  is 
inferred  that  the  first  has  to  the  last  of  the  first  rank  the  same 
ratio  which  the  first  of  the  other  rank  has  to  the  last.  This  is 
demonstrated  in  the  thirteenth  proposition  of  this  book. 

20.  Ex  cequo^  inversely :  when  there  are  three  or  more  mag- 
nitudes, and  as  many  others,  which  taken  two  and  two  in  a 
cross  order,  have  the  same  ratio ;  that  is,  when  the  first  magni- 
tude is  to  the  second  in  the  first  rank,  as  tlie  last  but  one  is  to 
the  last  in  the  second  rank;  and  the  second  to  the  third  of  the 
first  rank,  as  the  last  but  two  is  to  the  last  but  one  of  the  second 
rank,  and  so  on  ;  and  it  is  inferred,  as  in  the  preceding  defini- 
tion, that  the  first  is  to  the  last  of  the  first  rank,  as  the  first  to 
the  last  of  the  other  rank.  This  is  proved  in  the  fourteenth 
proposition  of  this  book. 


BOOK   IV.]  EUCLID   AND   LEGENDRE.  97 

AXIOMS. 

1.  Like  multiples  of  the  same,  or  of  equal  magnitudes,  are 
equal  to  one  another. 

2.  Those  magnitudes  of  which  the  same,  or  equal  magni- 
tudes, are  like  multiples,  are  equal  to  one  another. 

3.  A  multiple  of  a  greater  magnitude  is  greater  than  the 
same  multiple  of  a  less. 

4.  That  magnitude  of  which  a  multiple  is  greater  than  the 
same  multiple  of  another,  is  greater  than  that  other  magnitude. 

EXPLANATION    OF    SIGNS. 

1.  The  product  arising  from  multiplying  one  number  by  an- 
other is  expressed  by  writing  the  letters  representing  them,  one 
after  the  other,  without  any  sign  between  them ;  and  some- 
times by  placing  between  them  a  point,  or  the  sign  x. 

2.  A  product  is  called  a  power ^  when  tlie  factors  are  all  the 
same.  Thus,  AA,  or  as  it  is  generally  written,  A^,  is  called  the 
second  power,  or  the  square  of  A ;  AAA,  or  A',  the  third 
power,  or  cube  of  A;  AAA  A,  or  K\  its  fourth  power,  etc. 

In  relation  to  these  powers,  A  is  called  their  root.  Thus,  A 
is  the  second  or  square  root  of  A'^,  the  third  or  cube  root  of  A% 
the  fourth  root  of  A^,  etc.  In  like  manner,  the  second  or  square 
root  of  A  is  a  number  Avhich,  when  multiplied  by  itself,  pro- 
duces A;  the  th  rd  or  cube  ro  ^  of  A  is  such  a  number,  that  if 
it  be  multiplied  by  itself,  and  the  product  by  the  same  root 
again,  the  final  product  will  be  A.  The  square  root  of  A  is 
denoted  by  -/A  or  A*,  its  cube  root  by  -^A,  or  A*,  its  fourth 
root  by  Ai,  etc. 

3.  The  quotient  arising  from  dividing  one  number  by  an- 
other is  denoted  by  writing  the  dividend  as  the  numerator  of  a 
fraction,  and  the  divisor  as  its  denominator. 

4.  The  signs  =,  =0,  >,  <,  signify  respectively  equal  to^ 
equivalent  to,  greater  than,  less  than. 

PROPOSITIONS. 

Prop.  I. — Treor, — If  there  be  four  numbers  such  that  the 
quotients  obtained  by  dividing  the  first  by  the  second,  and  the 

1 


98  THE   ELEMENTS   OF  [bOOK  IV. 

third  hy  the  fourth,  are  equal ;  the  first  has  to  the  second  the 
same  ratio  that  the  third  has  to  the  fourth. 

Let  A,  B,  C,  D  be  four  magnitudes,  such  that  o— fj  >  ^^^^ 

A:  B::  C  :D. 

For,  let  m  and  n  be  any  whole  numbers,  and  multiply  the 

AC 

fractions  —  and  ^  by  m,  and  divide  the  product  by  n  y  then 

— =-  =  — ^-.     Now,  if  mA  be  srreater  than  riB,  mC  will  also  be 
nB     riD 

greater  than  nD ;  for,  if  this  were  not  so,  — rr  would  not  be 

equal  to  -y:-     In  like  manner  it  might  be  shown,  that  if  mA 

be  equal  to  7iB,  mC  will  be  equal  to  iiQ ;  and  that  if  raK  be 

less  than  nB,  rriG  will  be  less  than  wD.     But  wA,  raQ  are  any 

like  multiples  whatever  of  A,  C  ;  and  ?iB,  7^D  any  whatever  of 

B,  D  ;  and  therefore  (IV.  def  5)  A  :  B  : :  C  :  D.     Therefore,  if 

there  be  four  numbers,  etc. 

AC  1 

Scho.    This  proposition  is  the  same,  when  ^  or  Y\—P  ^^  ^iP 

being  a  whole  number. 

A     C 
Cor,  If  AD=BC,  by  dividing  by  B  and  D,  we  get  ^=t^> 

and  therefore,  by  this  proposition,  A  :  B  : :  C  :  D.  Hence,  if 
the  product  of  two  numbers  be  equal  to  that  of  two  others,  the 
one  pair  may  be  taken  as  the  extremes  and  the  other  as  the 
means  of  an  analogy. 

Prop.  II. — Theor. — If  any  four  numbers  he  proportional^ 

and  if  the  first  be  divided  by  the  second,  and  the  third  by  the 

fourth,  the  quotients  are  equal. 

A    C 
Let  A  :  B  : :  C  :  D ;  then  ^= jt 

If  A  and  B  be  whole  numbers,  let  the  first  and  third  terms 
be  multiplied  by  B,  and  the  second  and  fourth  by  A,  and  the 
products  are  AB,  AB,  BC,  AD.  Now,  since  the  first  and  sec- 
ond of  these  are  the  same,  the  third  and  fourth  are  (IV.  def.  6) 
equal;  that  is,  AD=BC;  and  by  dividing  these  by  B  and  D, 

A    C 

we  find  (IV.  ax,  2)  =t  =  ^. 


BOOK   IV.]  EUCLID   AND   LEGENDEE.  99 

If  A  and  B  be  fractions,  let  A=— ,  and  B=-,  so  that  niA— 

'  in  n 

E,  and  nB=F;  the  numerators  and  denominators  E,  F,  m,  n 

E    F 

beinii  whole  numbers.     Then  (hyp.)  — :  -  : :  C  :  D.      Multiply 

the  first  and  third  of  these  by  mF,  and  the  second  and  fourth 
l)y  wE,  and  the  products  are  EF,  EF,  mYQ,  and  nED.  Now, 
the  first  and  second  of  these  being  the  same,  the  third  and 
fourth  (IV.  def  5)  are  equal ;  that  is,  nED=mFC,  or  mnP^  — 
wiftBC,  since  E=imA,  and  F=wB.  Hence,  by  dividing  these 
by  m  and  w,  we  get  (IV.  ax.  2)  ADi=BC ;  and  the  rest  of  the 
proof  is  the  same  as  in  the  first  case.  Therefore,  if  any  four 
numbers,  etc. 

Scho.  1.  If  either  A  or  B  be  a  whole  number,  the  proof  is  in- 
cluded in  the  second  part  of  the  demonstration  given  above. 
Thus,  if  A  be  a  whole  number,  we  have  simply  E  =  A  and  m=: 
1,  and  everything  will  proceed  as  above.  The  proof  would  also 
be  readily  obtained  by  substituting  for  B  as  before,  but  retain- 
inc:  A  unchanged. 

If  A  and  B  be  incommensurable,  such  as  the  numbers  ex- 
pressing the  lengths  of  the  diagonal  and  side  of  a  square,  the 
lengths  of  the  diameter  and  circumference  of  a  circle,  etc.,  their 
ratios  may  be  approximated  as  nearly  as  we  please.  Thus  the 
diagonal  of  a  square  is  to  its  side,  as  |f  :  1,  nearly ;  as  \%\  :  1, 
more  nearly;  as  \%\\  :  1,  still  more  nearly,  etc.  Hence,  in 
such  cases  we  can  have  no  hesitation  in  admitting  the  truth  of 
the  proposition,  as  we  see  that  it  holds  with  respect  to  numbers 
the  ratio  of  which  differs  from  that  of  the  proposed  numbers  by 
a  quantity  which  may  be  rendered  as   small  as  we  please — 

smaller,  in  fact,  than  anything  that  can  be  assigned. 

A  1 

Scho.  2.  This  proposition  is  the  same,  when   T>=i?  or  -,    p 

being  a  whole  number. 

Scho.  3.  From  this  proposition  and  the  foregoing,  it  appears, 
that  if  two  fractions  be  equal,  the  numerator  of  the  one  is  to  its 
denominator  as  the  numerator  of  the  other  to  its  denominator ; 
and  that  if  the  first  and  second  of  four  proportional  numbers  be 
made  the  numerator  and  denominator  of  one  fraction,  and  the 
third  and  fourth  those  of  another,  the  two  fractions  are  equal. 


100  THE   ELEMENTS    OF  [bOOK   TV. 

This  is  the  same  in  substance  as  that  the  two  expressions,  A  : 

A     C 
B  : :  C  :  D,  and  |T=t-o  are   equivalent,  and  may  be  used  for 

one  another. 

Cor.  1.  It  appears  in  the  demonstration  of  this  proposition, 
that  AD  =  BC;  that  is,  if  four  numbers  be  pT'oportionals,  the 
product  of  the  extremes  is  equal  to  the  product  of  the  means. 
Hence,  if  the  product  of  the  means  be  di\ided  by  one  of  the  ex- 
tremes, the  quotient  is  the  other;  and  thus  we  have  a  proof  of 
the  ordinary  ai'ithmetical  rule  for  finding  a  fourth  proportional 
to  three  uiven  numV)ers. 

Cor.  2.  It  is  evident,  that  if  A  be  greater  than  B,  C  must  be 
greater  than  D  ;  if  equal,  equal ;  and  if  less,  less ;  as  otherwise 

=5"  and  =^  could  not  be  equal. 

A     C 
Cor.  3.  If  A  :  B  : :  C  :  D,  and  consequently —=y^,  by  multi- 

plying  these  fractions  by—,  we  get  — 7^= — :Fr,  or  mA  :  wB  :  : 

mC  :  nT>. 

A 

Cor.  4.  If  A  be  greater  than  B,  the  fraction  —  is  evidently 

T>  C  C 

greater  than  -^,  and  the  fraction  -^  less  than  ry ;  that  is,  of  two 

unequal  numbers,  the  greater  has  a  greater  ratio  to  a  third  than 
the  less  has ;  and  a  thii'd  number  has  a  greater  ratio  to  the 
less  than  it  has  to  the  greater. 

A  B 

Cor.  5.  Conversely,  if -^  be  greater  than -r^,  A  is  greater  than 

—  be  less  than  =, 
A  B' 


B ;  and,  if  —  be  less  than  z^,  A  is  also  greater  than  B. 


Prop.  III. — Theor. — If  four  numbers  he  proportionals,  they 
are  proportionals  also  when  taken  inversely. 

If  A  :  B  : :  C  :  D  ;  then,  inversely,  B  :  A  : :  D  :  C. 

For  (IV.  2,  cor.  1)  BC  =  AD ;  and  hence  by  dividing  by  A 

and  C,  we  obtain  -r=p5  or  (I^-  2,  scho.  2)  B  :  A  : :  D  :  C. 

Therefore,  if  four  numbers,  etc. 


/ 


BOOK  IV,]  EUCLID  AND  LEGENDRE.  101 

Prop.  IY. — Theor. — If  four  numbers  be  proportionals,  they 
are  also  proportionals  when  taken  alternately. 

If  A  :  B  : :  C  :  D ;  tlien,  alternately,  A  :  C  : :  B  :  D. 

For  (IV,  2,  cor.  1)  AD=;:BC  ;  whence,  by  dividing  by  C  and 

A     B 

D  we  get  7s  =yt;  or  (IV.  2,  scho.  2)  A  :  C  : :  B  :  D,     There- 

fore,  if  four  numbers,  etc, 

Scho.  When  the  first  and  second  terms  are  not  of  the  same 
kind  as  the  third  and  fourth,  the  terms  can  not  be  taken  altern- 
ately, as  ratios  would  thus  be  instituted  between  heterogene- 
ous macrnitudes. 


o 


Prop,  V. — Theor. — Ejual  numbers  have  the  same  ratio  to 
the  same  number  ^  and  the  same  has  the  same  ratio  to  equal 
numbers. 

Let  A  and  B  be  equal  numbers,  and  C  a  third ;  then  A :  C  : : 
B  :  C,  and  C  :  A  : :  C  :  B. 

A  B 

For,  A  and  B  being  equal,  the  fractions  -^  and   y,    are  also 

equal,  or,  which  is  the  same,  A  :  C  : :  B  :  C ;  and,  by  inversion, 
(IV.  3)  C  :  A  : :  C  :  B,     Therefore,  equal  numbers,  etc. 

Prop.  VI. — Theor. — JVmnbers  which  have  the  same  ratio 
to  the  same  nwnber  are  equal ;  and  those  to  which  the  same 
has  the  same  ratio  are  equal. 

If  A  :  C  : :  B  :  C,  or  if  C  :  A  : :  C  :  B,  A  is  equal  to  B. 

A     B 

For,  since  —7=.,  ,by  multiplying  by  C  we  get  A  =  B. 

The  proof  of  the  second  part  is  the  same  as  this,  since,  by  in- 
version (IV.  3),  the  second  analogy  becomes  the  same  as  the 
first.     Therefore,  numbers,  etc. 

Prop.  VII. — Theor. — Ratios  that  are  equal  to  the  sameratio 

are  equal  to  one  another. 

If  A  :  B  : :  C  :  D,  and  E  :  F  : :  C  :  D ;  then  A  :  B  : :  E :  F. 

-.        .        AC        ,  E     C    .       „       ,^  .  A     E      .    ^ 

ror,  smce  p=-r.,  and      =  -  ,therciore  (1.  ax.  1)  p=p, ;  that , 

is,  (IV.  2,  scho.  2)  A  :  B  : :  E  :  F.     Therefore,  ratios,  etc. 


102  THE   ELEMENTS   OF  [bOOK   17. 

Pkop.  Vin. — Theok. —  Ofnumhers  which  are  proportionals^ 
as  any  one  of  the  antecedents  is  to  its  consequent^  so  are  all  the 
antecedents  taken  together  to  all  the  conseqzients. 

If  A  :  B  : :  C  :  D  : :  E  :  F ;  then  A  :  B  : :  A+C+E  :  B+D 
+F. 

ACE 

Since  :^==:  ==-7,  put  each  fraction  equal  to  q,  and  multiply 

hj  the    denominators  ;   then    A=zBq,    C=:Dq,    and   E^Fg-. 
Hence,  by  addition,  A+C-}-E  =  (B  +  D-f  F)^'/  and  by  dividing 

by  B+D+F,  we  get  ^=WTWZw'     ^^^  ?— g  5  ^^^  there- 

foi'e  4"=t"!"^tS>  or  A  :  B  : :  A+C  +  E  :  B+D  +  F.     There- 
a    ±>+jj+i^ 

fore,  etc. 

Prop.  IX. — Theor. — 3fagmtudes  have  the  same  ratio  to  one 
another  that  their  like  multiples  have. 

Let  A  and  B  be  two  magnitudes ;  then,  n  being  a  whole 
number,  A  :  B  : :  nA  :  nB. 

For  :jj=— ^,  or  A  :  B  : :  nA  :  wB.  Therefore,  magnitudes, 
etc. 

Prop.  X. — ^Theor. — If  four  mimbers  be  prop>ortionals  ;  then 
(1)  hy  composition,  the  sum  of  the  first  and  seco7\d  is  to  the 
second,  as  the  sum  of  the  third  and  fourth  to  the  fourth  ;  and 
(2),  5y  division,  the  excess  of  the  first  above  the  second  is  to  the 
second,  as  the  excess  of  the  third  above  the  fourth  is  to  the 
fourth. 

If  A  :  B  : :  C  :  D  ;  then,  by  composition,  A+B  :  B  : :  C+D: 
D;  and  by  division,  A— B  :  B  : :  C — D  :  D. 

1.  Since  (hyp.)  vi  =  i-:i  and  since -rr  =  ^:  add  the  latter  frac- 

^  "^  ^  '  B     D  B     D 

tions  to  the  former,  each  to   each,  and  there  results  — rr~  = 

5^,  or  A+B  :  B  : :  C  +D  :  D. 

2.  By  subtracting  the  latter  pair  of  the  same  fractions  from 

the  former,  each  from  each,  we  obtain  — =—-  =  —=-—  ;  or  A — 

B  i) 

B  :  B  : :  C— D  :  D.     If,  therefore,  etc. 


BOOK  rV.]  EUCLID  AND  LEGENDRE.  103 

Cor.  By  dividing  the  fractions  which  were  found  above  by- 
addition,  by  those  which  were  found  by  subtraction,  we  get 

:^^=^^;  or  (IV.  2,  scho.  2)  A+B  :  A-B : :  C+D  :  C- 

D;  that  is,  if  four  numbers  be  proportional,  the  sura  of  the  fii-st 
and  second  terms  is  to  their  difference,  as  the  sum  of  the  third 
and  fourth  terms  is  to  their  difference.  It  is  evident,  that  if  B 
be  greater  than  A,  the  analogy  would  become  B+A  :  B— A  : : 
D  +  C  :  D— C. 

Prop.  XI. — Theoe. — If  foumumhers  he  proportional ;  then,, 
by  conversion,  the  first  is  to  its  excess  abuve  the  second^  as  the 
third  to  its  excess  above  the  fourth. 

If  A  :  B  : :  C  :  D ;  then,  by  conversion,  A  :  A — B  : :  C  : 
C— D. 

x^         •         n  ;,-  .BD  -.  AC. 

Jbor,  smce  (hyp.  and  mver.)  —=^^  and  smce  —  =  -;    take 

the  former  fractions  from  the  latter,  each  from  each,  and  there 

remains  — -r — = — ^ — ,  or  (by  inver.)  A:A  —  B::  C:C  — D. 

Therefore,  if  four  numbers,  etc. 

Prop.  XII. — Theor. — If  there  be  members  forming  two  or 
more  analogies  which  have  common  consequents^  the  sum  of 
all  the  first  antecedents  is  to  their  common  consequent,  as  the 
sum  of  all  the  other  antecedents  is  to  their  common  consequent. 

If  A  :  B  : :  C  :  D,  and  E  :  B  : :  F  :  D ;  then  A  +  E  :  .B  : : 
C+F  :  D. 

For  (hyp.)  -j5=— ,and  p  =  Tx;  and  hence,  by  addition,  — -^— 
=5^,  or  A+E  :  B  : :  C  +  F  :  D.     If,  therefore,  etc. 

Prop.  XIII. — Theor. — If  there  be  three  or  more  niiyn^ers, 
and  OS  many  others,  which,  taken  two  and  two  in  order,  have 
the  sa7ne  ratio  ;  then,  ex  aequo,  the  first  has  to  the  last  of  the 
first  rank  the  sayne  ratio  that  the  first  has  to  the  last  of  the 
second  rank. 

If  the  two  ranks  of  numbers.  A,  B,  C,  D,  and  E,  F,  G,  II,  be 


104  THE   ELKMENT8    OF  [bOOK    IV. 

Buch  that  A  :  B  : :  E  :  F,  B  :  C  : :  F  :  G,  and  C  :  D  : :  G  :  11 ; 
then  A  :  D  : :  E  :  H. 

A     E  B    F  C     C 

For,  since  (hyp.)  --  =-,  -  =  -  and  ^  =  - ;  by  multiplying 

together  the  first,  third,  and  fifth  fractions,  and  the  second, 
fourth,  and  sixth,  we  obtain  jTnn^KnxT  5  ^r,  by  dividing  the 

terras  of  the  first  of  these  fractions  by  BC,  and  those  of  the 

A    E 

second  by  FG,  j)=g,  or  A  :  D  : :  E  :  II.     Therefore,  if  there 

be  three,  etc. 

This  proposition  might  also  be  enunciated  tlius  :  If  there  be 
numbers  forming  two  or  more  analogies,  such  that  the  conse- 
quents in  each  are  the  antecedents  in  the  one  immediately  fol- 
lowing it,  an  analogy  will  be  obtained  by  taking  the  antece- 
dents of  the  first  analogy  and  the  consequents  to  the  last  for 
its  antecedents  and  consequents. 

Prop.  XIV. — Theok. — If  there  be  three  or  more  numbers, 
and  as  many  others,  which,  taken  two  and  two  in  a  cross  order, 
have  the  same  ratio ;  then,  ex  tequo  inversely,  the  first  has  to 
the  last  of  the  first  rank  the  same  ratio  uhlch  the  first  has  to 
the  last  of  the  second  ranTc. 

If  the  two  ranks  of  numbers.  A,  B,  C,  D,  and  E,  F,  G,  H,be 
such  that  A  :  B  : :  G  :  H,  B  :  C  : :  F  :  G,  and  C  :  D  : :  E  :  F ; 
then,  ex  mqrco  inversely,  A  :  D  : :  E  :  H. 

A     C   B     F  P      F 

For,  since  (hyp.)  -^=^,  ^=^,and  ^  =^,  by  multiplying  to- 
gether the  fractions  as  in  the  preceding  proposition,  we  get 

ABC     GFE        ,  ,,..,.,  .    ,      .      ^ 

^pY)=TT7TT^j  whence,  by  dividing  the  terms  of  the  first  of 

these  fractions  by  BC,  and  those  of  the  second  by  GF,  we  ob- 

A     E 
tain  jx  =  fT,  or  A  :  D  : :  E  :  H.     If,  therefore,  etc. 

This  proposition  may  also  be  enunciated  thus:  If  there  be 
numbers  forming  two  or  more  analogies,  such  that  the  means 
of  each  are  the  extremes  of  the  one  immediatelj'  following  it, 
another  analogy  may  be  obtained  by  taking  the  extremes  of 
the  first  analogy  and  the  means  of  the  last  for  its  extremes  and 
means. 


BOOK  IV.]         EUCLID  AND  LEGENDKE.  105 

Prop.  XV. — Theor. — If  there  he  numbers  forming  tico  or 
more  analogies^  the  products  of  their  corresponding  terms  are 
proportionals. 

If  A  :  B  : :  C  :  D,  E  :  F  : :  G  :  II,  and  K  :  L  : :  M  :  N ;  then 
AEK  :  BFL  : :  CGM  :  DUN. 

^      ,,       -  A     C    E     G  TC     M        ^      .  . 

For  (hyp.)  j^  =0,  ^=|j>'  aucl  jy  =  ^;  and  taking  the  pro- 
ducts of  the  corresponduig  terms  of  these  fractions,  we  obtain 

A  FTC     OC  \f 

^^=^^.,  or  AEK  :  BFL  ::  CGM  :  DHN.     Therefore,  if 

Bi'L     DHIS' 

there  be  numbers,  etc. 

Cor.  1.  Hence,  if  there  be  two  analoj^ies  consistinsr  of  the 
sane  terms,  A,  B,  C,  D,  we  have  A" :  B'  : :  C^ ;  D^ ;  if  there  be 
three,  we  have  A'  :  B'  : :  C*  :  D',  etc. ;  and  it  thus  appears, 
that  like  powers  of  proportional  numbers  are  themselves  pro- 
portional. 

Cor.  2.  Like  roots  of  proportional  numbers  are  proportional. 
Thus,  if  A  :  B  : :  C  :  D,  let  4/ A  :  VB  : :  VC  :  VE.  Then,  by 
the  preceding  corollary,  A  :  B  : :  C  :  E.  But  (hyp.)  A  :  B  : : 
C  :  D  ;  and  therefore  (IV.  7)  C  :  E  : :  C  :  D,  and  (IV.  6)  E  = 
D,  and  consequently  V  :  A  -/  B  : :  VC  :  VE,  or  VD. 

Prob.  XVI. — Theor. — The  sum  of  the  greatest  and  least  of 
four  p.  oportional  members  is  greater  than  the  sum  of  the  other 
two. 

If  A  :  B  : :  C  :  D,  and  if  A  be  the  greatest,  and  therefore 
(IV.  2,  cor.  2)  D  the  least ;  A  and  D  are  together  greater  than 
B  and  C. 

For  (by  conversion)  A  :  A — B  ::  C  :  C — D,  and,  altern- 
ately, A  :  C  : :  A— B  :  C— D.  But  (hyp.)  A>C,  and  there- 
fore'(IV.  2,  cor.  2)  A— B  >  C  — D.  To  each  of  these  add  B  ; 
then  A>B  +  C— D.  Add  again,  D;  then,  A-hD>B4-C. 
Therefore,  etc. 

Cor.  Hence  the  mean  of  three  proportional  numbers  is  less 
than  half  the  sum  of  the  extremes. 

Prob.  XVII. — Theor. — In  numbers  which  are  continual 
proportiojials,  the  first  is  to  the  third  as  the  second  power  of 
the  first  to  the  second  power  of  the  second  y  the  first   to  the 


106  THE   ELEMENTS   OF  [bOOK   IV. 

fourth  as  the  third  power  of  the  first  to  the  third  power  of  the 
second  ;  the  first  to  the  fifth  as  the  fourth  power  of  the  first  to 
the  fourth  power  of  the  second  ;  and  so  on. 
1      If  A,  Vy,  C,  D,  E,  etc.,  be  continual  proportionals;  A  :  C  :  : 
A^  B^ ;  A  :  D  :  :  A^ :  B';  A  :  E  :  :  A^  :  B',  etc. 

For,  since  (IV.  def.  8)  A  :  B  :  :  B  :  C,  and  since  A  :  B  : : 
A  :  B,  we  have  (IV.  15)  A^ :  B^  : :  AB  :  BC,  or,  dividing  the 
third  and  fourth  terms  by  B,  A*  :  B'  :  :  A  :  C. 
Again :  since  A* :  B' : :  A  :  C, 

and  A  :  B  :  :  C  :  D  we  have  (IV.  15) 
A'  :  B'  : :  AC  :  CD,  or  dividing  the  third  and  fourth  terms 
by  C,  A^ :  B^ : :  A  :  D  ;  and  so  on,  as  far  as  we  please.  There- 
foi'e,  etc. 

Cor.  Hence  (IV.  defs.  11  and  12)  the  ratio  which  is  duplicate 
of  that  of  any  two  numbers,  is  the  same  as  the  ratio  of  their 
squares ;  that  which  is  triplicate  of  their  ratio,  the  same  as  the 
ratio  of  their  cubes,  etc. 

Prop,  XVIII. — ^Theor. — A  ratio  xcMch  is  compounded  of 
other  ratios,  is  the  same  as  the  ratio  of  the  products  of  their 
homologous  terms. 

Let  the  ratio  of  A  to  D  be  compounded  of  the  ratios  of  A  to 
B,  B  to  C,  and  C  to  D ;  the  ratio  of  A  to  D  is  tlie  same  as  that 
of  ABC,  the  product  of  the  antecedents,  to  BCD,  the  product 
of  the  consequents. 

For,  since  A  :  D  : :  A  :  D,  multiply  the  terms  of  the  sec- 
ond ratio  by  BC  ;  then  (IV.  9)  A  :  D  : :  ABC  :  BCD.  There- 
fore, etc. 

Prop.  XIX. — Theor. — In  numbers  which  are  continual 
pro2)ortionalsj  the  difference  of  the  first  and  second  is  to  the 
first,  as  the  difference  of  the  first  and  last  is  to  the  sum  of  all 
t/ie  terms  excejjt  the  last. 

If  A,  B,  C,  D,  E  be  continual  proportionals,  A — B  :  A  : : 
A— E:  A  +  B  +  C  +  D. 

For,  since  (hyp.)  A  :  B  : :  B  :  C  :  :  C  :  D  : :  D  :  E,  we 
have  (IV.  8)  A  :  B 


(conv.)     A  :  A — B  : 
A— B  :  A  ::  A— E 


A+B+C+D  :  B  +  C  +  D  +  E.  Hence 
A  +  B  +  C  +  D  :  A— E;  and  (inver.) 
A+B  +  C  +  I). 


BOOK  IV.]         EUCLID  AND  LEGENDRE.  107 

It  is  evident  that  if  A  were  the  least  term,  and  E  the  great- 
est, we  should  get  in  a  similar  manner,  B^ — A  :  A  : :  E — A  : 
A-f-B-f-C+D.     Therefore,  in  numbers,  etc. 

Cot.  If  the  series  be  an  infinite  decreasing  one,  the  last  term 
will  vanish,  and  if  S  be  put  to  denote  the  sum  of  the  series,  the 
analogy  will  become  A — B  :  A  : :  A  :  S ;  and  this,  if  rA  be 
put  instead  of  B,  and  the  first  and  second  terms  be  divided  by 
A,  will  be  changed  into  1 — r  :  1  : :  A  :  S.  The  number  r  is 
called  the  common  ratio,  or  common  m,ultiplier,  of  the  series, 
as  by  multiplying  any  term  by  it,  the  succeeding  one  is  ob- 
tained. 


END    OP   BOOK   FOURTH. 


BOOK    FIFTH. 

DEFINITIONS. 

1.  Similar  rectilineal  figures  are  those  which  have  their 
several  angles  equal,  each  to  each,  and  the  sides  about  the 
equal  angles  proportionals. 

2.  Two  magnitudes  are  said  to  be  reciprocally  proportional 
to  two  others,  when  one  of  the  first  pair  is  to  one  of  the  second, 
as  the  remainincc  one  of  the  second  is  to  the  remaining  one  of 
the  first. 

3.  A  straight  line  is  said  to  be  cut  in  extreme  andmean  ratio^ 
■when  the  whole  is  to  one  of  the  segments  as  that  segment  is  to 
the  other. 

4.  The  altitude  of  any  figure  is  the  straight  line  drawn  from 
its  vertex  perpendicular  to  its  base. 

5.  A  sti-aight  line  is  said  to  be  cut  harmonically,  Avhen  it  is 
divided  into  three  segments,  such  that  the  whole  line  is  to  one 
of  the  extreme  segments  as  the  other  extreme  segment  is  to  the 
middle  one. 

PROPOSITIONS. 

Prop.  I. — Theor. — Triangles  and  parallelograms  of  the 
same  altitude  are  one  to  another  as  their  bases.         ^ 

Let  the  triangles  ABC,  ACD,  and  the  parallelograms  EC, 
CF  have  the  same  altitude,  viz.,  the  perpendicular  drawn  from 
the  point  A  to  BD ;  then,  as  the  base  BC  is  to  the  base  CD,  so 
is  the  triangle  ARC  to  the  triangle  ACD,  and  the  parallelo- 
gram EC  to  the  parallelogram  CF. 

Produce  BD  both  ways,  and  take  any  number  of  straight 
lines  BG,  Gil,  each  equal  to  BC  ;  and  any  number  DK,  KL, 
each  equal  to  CD;  and  join  AG,  AH,  AK,  AL.  Then,  because 
CB,  BG,  Gil  are  all  equal,  the  triangles  ABC,  AGB,  AUG  are 
(L  15,  cor.)  all  equal.     Therefore,  whatever  multiple  the  base 


BOOK   v.] 


EUCLID   AND    LEGENDRE. 


loa 


HC  is  of  BC,  the  same  multiple  is  the  triangle  AUG  of  ABC. 
For  the  same  reason,  wliatver  multiple  LC  is  of  CD,  the  same 
multiple   is  the  triangle  ALC 
of  ADC.    Also,  if  the'base  HC  - 
be  equal   to   CL,   the  triangle 
AHC  is  equal  (I,   15,  cor.)  to 
ALC  ;  and  if  the  base  HC  be 
greater  than  CL,   likewise  (L 
15,  cor.  6)  the  ti-iangle  AHC  is 

greater  than  ALC;  and  if  less,  less.  Therefore,  since  there  are 
four  magnitudes,  viz.,  the  two  bases,  BC,  CD,  and  the  two  tri- 
angles ABC,  ACD;  and  of  the  base  BC,  and  the  triangle  ABC, 
the  first  and  third,  any  like  multiples  whatever  have  been  taken, 
viz.,  the  base  HC,  and  the  triangle  AHC ;  and  of  the  base  CD, 
and  the  triangle  ACD,  the  second  and  fourth,  have  been  taken 
any  like  multiples  whatever,  viz.,  the  base  CL,  and  the  triangle 
ALC  ;  and  that  it  has  been  shown  that,  if  the  base  HC  be 
greater  than  CL,  the  triangle  AHC  is  greater  than  ALC ;  if 
equal,  equal;  and  if  less,  less;  therefore  (IV.  def.  5)  as  the 
base  BC  is  to  the  base  CD,  so  is  the  triangle  ABC  to  the  trian- 
gle ACD. 

Again  :  because  (L  15,  cor.)  the  parallelogram  CE  is  double 
of  the  triangle  ABC,  and  the  parallelooi-ani  CP""  of  the  triangle 
ACD,  and  that  (IV.  9)  magnitudes  have  the  same  ratio  which 
their  like  multiples  have ;  as  the  triangle  ABC  is  to  the  trian- 
gle ACD,  so  is  the  parallelogram  EC  to  the  parallelogram  CF. 
But  it  has  been  shovvn,  that  BC  is  to  CD,  as  the  triangle  ABC 
to  the  triangle  ACD ;  and  as  the  tiiansjfle  ABC  is  to  the  trian- 
gle  ACD,  so  is  the  parallelogram  EC  to  the  parallelogram  CF; 
therefore  (TV.  7)  as  the  base  BC  is  to  the  base  CD,  so  is  the 
parallelogram  EC  to  the  parallelogram  CF.  Wheiefore,  trian- 
gles, etc. 

Scho.  This  proposition  may  be  briefly  demonstiated  thus: 

Let  a  perpendicular  drawn  from  A  to  BD  be  called  P.     Then, 

■J^P.BC  will  be  equivalent  to  the  area  of  the  triangle  ABC,  and 

Ap.CD  that  of  ACD.     Dividing,  therefore,  the  former  of  these 

,    t,      ,     1  il'-KC         BC     ABC         ,„,    „ 

equals  by  the  latter,  we  get  yp-^  or,  qy)~A<Jd'  °^"  ^  ' 

scbo.  2)  BC  :  CD  : :  ABC  :  ACD.     In  extending  this  method 


110  THE   ELEMENTS   OF  [BOOK  V. 

of  proof  to  the  parallelograms,  we  have  merely  to  hbc  P  instead 
ofiP. 

Cor.  1.  From  this  it  is  plain,  that  triangles  and  parallelo- 
grams which  have  equal  altitudes,  are  one  to  another  as  their 
bases. 

Let  the  figures  be  placed  so  as  to  have  their  bases  in  the 
same  straight  line ;  and  perpendiculars  being  drawn  from  the 
vertices  of  the  triangles  to  the  bases,  the  straight  line  Avhich 
joins  the  vertices  is  parallel  (I,  15,  cor.)  to  that  in  Avhich  their 
bases  are,  because  the  perpendiculars  ai-e  both  equal  and  paral- 
lel to  one  another.  Then,  if  the  same  construction  be  made  as 
in  the  proposition,  the  demonstration  will  be  the  same. 

Cor.  2.  Hence,  if  A,  B,  C  be  any  three  straight  lines,  we  have 
A :  B  : :  A.C  :  B.C. 

Cor.  3.  So,  likewise,  if  the  straight  lines  A,  B,  C,  D  be  pro- 
portional, and  E  and  F  be  any  other  straight  lines,  we  shall 
have,  according  to  the  preceding  corollary,  and  the  seventh 
proposition  of  the  fourth  book,  A.E  :  BE  : :  C.F  ;  D.F. 


Prop.  II. — Theor. — If  a  straight  line  he  parallel  to  the  base 
of  a  triangle,  it  cuts  the  other  sides,  or  those  produced,  propor- 
tionally, and  the  segments  between  the  parallel  and  the  base  are 
homol'jgous  to  one  another ;  and  (2)  if  the  sides  of  a  triangle, 
or  the  sides  produced,  be  cut  proportionally,  so  that  the  seg- 
ments between  the  points  of  section  and  the  base  are  homologous 
to  one  another,  the  straight  line  which  jo.  ns  the  points  of  sec- 
tion is  parallel  to  the  base. 

The  enunciation  of  this  proposition  which  is  given  by  Dr. 
Simson  and  others,  is  defective,  and  might  lead  to  error  in 
its  application,  as  it  does  not  point  out  what  lines  are  homolo- 
gous to  one  another  in  the  analogies. 

It  is  plain  that,  instead  of  one  proposition,  this  is  in  reality 
two,  which  are  converses  of  one  another. 

1.  Let  DE  be  parallel  to  BC,  one  of  the  sides  of  the  triangle 
ABC  ;  BD  :  DA  : :  CE  :  EA. 

Join  BE,  CD.  Then  (L  15,  cor.)  the  triangles  BDE,  CDE 
are  equivalent,  because  they  are  on  the  same  base  DE,  and  be- 
tween the  same  parallels.  DE,  BC.     Now  ADE  is  another  tri- 


BOOK   v.]  EUCLID    AND   LEGENDRK  111 

angle,  and  (IV.  5)  equal  magnitudes  have  to  the  same  the  same 
ratio;  therefore,  as  the  triangle 

BDE  to  ADE,  so  is  the  triangle         ^  ^  » 

CDE  to  ADE.     But, 

(V.  ])  as  the  triangle  BDE 
to  ADE,  so  is  BD  to  DA ; 

Because,  having  the  same  alti- 
tude, viz.,  the  perpendicular 
drawn  from  E  to  AB,  they  are 
to  one  another  as  their   bases;         ^  c         B  O 

and  for  the  same  reason, 

as  the  triangle  CDE  to  ADE,  so  is  CE  to  EA. 

Therefore  (IV.  1)  as  BD  :  DA  : :  CE  :  E A. 

2.  Next,  let  the  sides  AB,  AC  of  the  triangle  ABC,  or  those 
produced,  be  cut  proportionally  in  the  points  D,  E  ;  that  is,  so 
that  BD  :  DA  : :  CE  :  EA,  and  join  DE  ;  DE  is  parallel  to  BC. 

The  same  construction  being  made,  because  (hyp.) 

as  BD  :  DA  : :  CE  :  EA  ;  and  (V.  1) 

as  BD  to  DA,  so  is  the  triangle  BDE  to  the  triangle  ADE;  and 
as  CE  to  EA,  so  is  the  triangle  CDE  to  ADE ;  therefore  (IV. 
7)  the  triangle  BDE  is  to  ADE,  as  the  triangle  CDE  to  ADE ; 
that  is,  the  triangles  BDE,  CDE  have  the  same  ratio  to  ADE ; 
and  therefore  (IV.  6)  the  triangles  BDE,  CDE  are  equal ;  and 
they  are  on  the  same  base  DE,  and  on  the  same  side  of  it ; 
therefore  (I.  15,  cor.)  DE  is  parallel  to  BC.  Wherefore,  if  a 
straight  line,  etc. 

Cor.  The  triangles  which  two  intersecting  straight  lines 
form  with  two  parallel  ones,  have  their  sides  which  are  on  the 
intersecting  lines  proportional ;  and  those  sides  are  homologous 
which  are  in  the  same  straight  line;  and  (2),  conversely,  if  two 
straight  lines  form  with  two  intersecting  ones  triangles  which 
have  their  sides  that  are  on  the  intersecting  lines  proportional, 
the  sides  which  are  in  the  same  Rtrai<irht  line  with  one  another 
being  homologous,  those  straight  lines  ai-e  parallel. 

1.  Let  DE  and  BC  (first  and  second  figures)  be  the  parallels, 
and  let  them  be  cut  by  the  straight  lines  BD,  CE,  which  inter- 
sect each  other  in  A;  then  BA  :  AC  : :  DA  :  AE.  For,  since 
BD  :  DA  : :  CE  :  EA,  we  have,  by  composition  in  the  first  fig- 


112  THE   ELEMENTS    OF  [BOOK   V. 

tire,  and  by  division  in  the  second,  BA  :  DA  : :  CA  :  EA,  and, 
alternately,  BA  :  AC  : :  DA  :  AE. 

2.  But  if  BA  :  AC  :  :  DA  :  AE,  DE  and  BC  are  parallel. 
For,  alternately,  BA  :  DA  ::  CA  :  EA  ;  then,  in  the  first  figure 
by  division,  and  in  the  second  by  composition,  we  have  BD  : 
DA  : :  CE  :  EA  ;  and  therefore,  by  the  second  part  of  this 
proposition,  DE  is  pai-allel  to  BC. 

Prop.  III. — TheoPw — Tlie  sides  about  the  equal  aiigles  of 
equiangular  triangles  are  proportionals  ;  and  those  ichich  are 
ojyposite  to  the  equal  angles  are  homologous  sides^  that  is,  are 
the  antecedents  or  consequents  of  the  nitios. 

Let  ABC,  DCE  be  equiangidar  triangles,  having  the  angle 
ABC  equal  to  DCE,  and  ACB  to  DEC,  and  consequently  (I. 
20,  cor.  5)  BAC  equal  to  CDE ;  the  sides  about  the  equal 
angles  are  proportionals ;  and  those  are  the  homologous  sides 
which  are  opposite  to  the  equal  angles. 

Let  the  triangles  be  placed  on  the  same  side  of  a  straight 
line  BE,  so  that  sides  BC,  CE,  which  are  opposite  to  equal 
ano-les,  may  be  in  that  straiu^ht  line  and  contii^nous  to  one  an- 
other;  and  so  that  neither  the  equal  angles  ABC,  DCE,  nor 
ACB,  DEC  at  the  extremities  of  those  sides  may  be  adjacent. 

Then,  because  (I.  20)  the  angles 
ABC,  ACB  are  together  less  than 
two  right  angles,  ABC  and  DEC, 
which  (hyp.)  is  equal  to  ACB,  are 
also  less  than  two  right  angles ; 
wherefore  (I.  19)  BA,  ED  will  meet, 
if  produced ;  let  them  be  produced 
and  meet  in  F.  Again :  because 
the  angle  ABC  is  equal  to  DCE,  BF  is  parallel  (L  16,  cor.)  to 
CD  ;  and,  because  the  angle  ACB  is  equal  to  DEC,  AC  is  par- 
allel to  FE.  Therefore,  FACD  is  (I.  def  15)  a  parallelogram; 
and  consequently  (L  15,  cor.)  AF  is  equal  to  CD,  and  AC  to 
FD.  Now  (V.  2)  because  AC  is  parallel  to  FE,  one  of  the 
sides  of  the  triangle  FBE, 

BA  :  AF  : :  BC  :  CE. 
But  AF  is  equal  to  CD ;  therefore  (IV.  5) 

as  BA  :  CD  : :  BC  :  CE, 


BOOK  v.]  EUCLID  AND  LI  GKNDRE.  113 

and  alternately  (IV.  4)  as  AB  :  BC  : :  DC  :  CE. 
Ai^ain :   (V.  2)  because  CD  is  parallel  to  15F,  as 

BC  :  CE  : :  FD  :  DE;  but  FD  is  equal  to  AC;  therefore, 
as  BC  :  CE  : :  AC  :  DE  ;  and,  alternately, 
as  BC  :  CA  : :  CE  :  ED. 
Therefore,  because  it  has  been  proved  that 

AB  :  BC  : :  DC  :  CE,  and  as  BC  :  CA  : :  CE  :  ED; 
ex  mquo    (IV.    13),   BA  :  AC  : :  CD  :   DE.     Therefore,   the 
Bides,  etc. 

Scho.  1.  Hence  (V,  def  l)  equianirnlar  triangles  are  similar. 

Cor.  If  two  angles  of  one  triangle  be  respectively  equal  to 
two  angles  of  another,  their  sides  are  proportional,  and  the 
sides  opposite  to  equal  angles  are  homologous.  For  (I.  20,  cor. 
5)  the  remaining  angles  are  equal,  and  therefore  the  triangles 
are  equiangular. 

Scho.  2.  In  a  similar  manner  we  may  produce  a  given 
straight  line,  so  that  the  whole  line  so  produced  may  have  to 
the  part  produced  the  ratio  of  two  given  straight  lines.  Thus, 
if  BA  be  the  line  to  be  produced,  make  at  B  an  angle  of  any 
magnitude,  and  take  BE  and  CE  equal  to  the  other  given  lines; 
join  AC,  and  draw  FE  parallel  to  it.  Then,  since  FE  is  paral- 
lel to  AC,  a  side  of  the  triangle  ABC,  we  have  (V.  2)  BF  : 
AF  : :  BE  :  CE,  so  that  BF  has  to  AF  the  given  ratio. 

Prop.  IV. — Theor. —  The  straight  line  which  bisects  an  aru- 
gle  of  a  triangle,  divides  the  opposite  side  into  segments  ichich 
have  the  same  ratio  to  one  another  as  the  adjacent  sides  of  the 
triangle  have ;  and  (2)  if  the  segm.ents  of  the  base  have  the 
same  ratio  as  the  adjacent  sides,  the  straight  line  draicn  from 
the  vertex  to  the  point  of  section,  bisects  the  vertical  angle. 

1.  Let  the  angle  BAC  of  the  triangle  ABC  be  bisected  by 
the  straight  line  AD  ;  then  BD  :  DC  : :  BA  :  AC. 

Through  C  draw  (I.  18)  CE  par- 
allel to  DA;  then  (I.  16,  cor.  l) 
BA  produced  will  meet  CE;  let 
them  meet  in  E.  Because  AC 
meets  the   parallels  AD,  EC,  the      g  D        c 

angle  ACE  is  equal  (I.  16)  to  the 

alternate  angle  CAD  ;  and  because  BAE  meets  the  same  paral- 
8 


114 


THE   ELEMENTS   OF 


[book 


lels,  the  angle  E  is  equal  (I.  ]  6,  part  2)  to  BAD ;  therefore  (I. 
ax.  1)  the  angles  ACE,  AEC  are  equal,  because  they  are  re- 
spectively equal  to  the  equal  angles,  DAC,  DAB;  and  conse- 
quently AE  is  equal  (I.  1,  cor.)  to  AC.  Now  (V.  2)  because 
AD  is  parallel  to  EC,  one  of  the  sides  of  the  triangle  BCE, 
BD  :  DC  : :  BA  :  AE ;  but  AE  is  equal  to  AC;  therefore  (IV. 
5)  BD  :  DC  : :  BA  :  AC. 

2.  Let  now  BD  :  DC  : :  BA  :  AC,  and  join  AD ;  the  angle 
BAC  is  bisected  by  AD. 

The  same  construction  being  made,  because 
(hyp.)  BD  :  DC  : :  BA  :  AC;  and 
(V.  2)  BD  :  DC  : :  BA  :  AE, 
since  AD  is  parallel  to  EC  ;  therefore  (IV.  1)  BA  :  AC  : :  BA  : 
AE;  consequently  (IV.  6)  AC  is  equal  to  AE ;  and  (I.  1)  the 
angles  AEC,  ACE  are  therefore  equal.     But  (I.  16)  the  angle 
BAD  is  equal  to  E,  and  DAC  to  ACE ;  wherefore,  also,  BAD 
is  equal  (I.  ax.  1)   to  DAC;  and  therefore  the  angle  BAC  is 
bisected  by  AD.     The  straight  line,  therefore,  etc. 

And  if  an  exterior  angle  of  a  triangle  be  bisected  by  a  straight 
line  which  also  cuts  the  base  produced,  the  segments  between 
the  bisecting  line  and  the  extremities  of  the  base  have  the 
same  ratio  to  one  another  as  the  other  sides  of  the  triangle 
have ;  and  (2)  if  the  segments  of  the  base  produced  have  the 
same  ratio  which  the  other  sides  of  the  triangle  have,  the 
straight  line  drawn  fiom  the  vertex  to  the  point  of  section 
bisects  the  exterior  angle  of  the  triangle. 

1.  Let  an  exterior  angle  CAE  of  any  triangle  ABC  be 
bisected  by  AD  which  meets  the  opposite  side  produced  in  D  ; 
then  BD:  DC::  BA  :  AC. 

Through  C  draw  (T.  18)  CF  parallel  to  AD;  and  because 
AC  meets  the  parallels  AD,  FC,  the  angle  ACF  is  equal  (I.  16) 

to  CAD ;  and  because  the  straight  line 
FAE  meets  the  parallels  AD,  FC,  the 
angle  CFA  is  equal  to  DAE ;  therefore, 
also,  ACF",  CFA  are  (I.  ax.  1 )  equal  to 
one  another,  because  they  are  respect- 
ively equal  to  the  equal  angles  DAC, 
DAE  ;  and  consequently  (I.  cor.)  AF  is 
equal  to  AC.     Then  (V.  2)  because  AD  is  parallel  to  FC,  a  side 


BOOK  v.]  EUCLID  AND  LEGENDEE.  115 

of  the  triangle  BCF,  BD  :  DC  : :  BA  :  AF ;  but  AF  is  equal  to 
AC ;  as  therefore  BD  :  DC  : :  BA  :  AC. 

2.  Let  now  BD  :  DC  : :  BA  :  AC,  and  join  AD ;  the  angle 
CAD  is  equal  to  DAE. 

The  same  construction  being  made,  because 
BD  :  DC  : :  BA  :  AC;  and  that  (V.  2)  BD  :  DC  : :  BA  :  AF; 
therefore  (IV.  V)  BA  :  AC  : :  BA  :  AF  ;  wherefore  (IV.  6)  AC 
is  equal  to  AF,  and  (I.  1)  the  angle  AFC  to  ACF.  But  (I.  16) 
the  angle  AFC  is  equal  to  EAD,  and  ACF  to  CAD ;  therefore, 
also  (I.  ax.  1),  EAD  is  equal  to  CAD.     Wherefore,  etc. 

Cor.  If  G  be  the  point  in  which  BC  is  cut  by  the  straight 
line  bisecting  the  angle  BAC, 

we  have  (V.  4)  BG  :  GC  :  :  BA  :  AC ; 
and  by  this  proposition,  BD  :  DC  : :  BA  :  AC ; 
whence  (IV.  7)  BD  :  DC  ::  BG  :  GC,  and  therefore  (V.  def  5) 
BD  is  divided  harmonically  in  G  and  C. 

8cho.  If  the  triangles  be  isosceles,  the  line  bisecting  the  ex- 
terior angle  at  the  vertex  is  parallel  to  the  base.  In  this  case, 
the  segments  may  be  regarded  as  infinite,  and  therefore  equal, 
their  difference,  the  base,  being  infinitely  small  in  comparison 
of  them. 

Prop.  V. — Theor. — If  the  sides  of  two  triangles^  about  each 
of  their  angles^  he  proportionals,  the  triangles  are  equiangular, 
and  have  their  equal  angles  opposite  to  the  homologous  sides. 

Let  the  triangles  ABC,  DEF  have  their  sides  proportionals, 
that  AB  :  BC  :  I  DE  :  EF ;  and  BG  :  CA  : :  EF  :  FD ;  and 
consequently,  ex  aequo,  BA  :  AC  : :  ED  :  DF;  the  triangles  are 
equiangular,  and  the  equal  angles  are  opposite  to  the  homolo- 
gous sides,  viz.,  the  angle  ABC  equal  to  DEF,  BCA  to  EFD, 
and  BAC  to  EDF. 

At  the  points  E,  F,  in  the  straight 
line  EF,  make  (L  13)  the  angle  FEG 
equal  to  B,  and  EFG  equal  to  C. 
Then  (V.  3,  cor.)  the  triangles  ABC, 
GEF  have  their  sides  opposite  to  the 
equal  angles  proportionals  ;  wherefore, 

AB:BC::GF:EF;  but  (hyp.) 
AB  :  BC  : :  DE  ;  EF. 


116  THE   ELEMENTS   OF  [bOOK  V. 

Therefore  (TV.  7)  DE  :  EF  : :  GF  :  EF ;  whence,  since  DE  and 
GF  have  the  same  ratio  to  EF,  they  are  (IV.  6)  equal.  It 
may  be  shown  in  a  similar  manner  that  DF  is  equal  to  EG; 
ani  because,  in  the  triangles  DEF,  GEF,  DE  is  equal  to  FG, 
EF  common,  and  DF  equal  to  GE ;  therefore  (I.  4)  the  angle 
DEF  is  equal  to  GFE,  DFE  to  GEF,  and  EDF  to  EGF.  Then, 
because  the  angle  DEF  is  equal  to  GFE,  and  (const.)  GFE  to 
ABC  ;  therefore  the  angle  ABC  is  equal  to  DEF.  For  the 
same  reason,  ACB  is  equal  to  DFE,  and  A  to  D.  Therefore 
the  ti'iangles  ABC,  DEF  are  equiangular.  Wherefore,  if  the 
sides,  etc. 

Prop.  VI.— Theor. — If  two  triangles  have  one  angle  of  the 
one  equal  to  one  angle  of  the  other^  and  the  sides  about  the 
equal  angles  proportionals ;  the  remaining  angles  are  equal^ 
each  to  each^  viz.y  those  which  are  opposite  to  the  homologous 
sides. 

Let  the  triangles  ABC,  GEF,  of  the  previous  diagrams,  have 
the  angles  ABC,  EFG  equal,  and  the  sides  about  those  angles 
proportionals ;  that  is,  BA  :  BC  :  :  GF  :  EF ;  the  angle  BAG 
is  equal  to  EGF,  and  ACB  to  EFG. 

Make  (I.  13)  the  angle  FED  equal  to  either  of  the  angles 
ABC,  GFE ;  and  the  angle  EFD  equal  to  ACB.  Then  (V.  3, 
cor.), 

BA  :  BC  : :  DE  :  EF.     But  (hyp.) 
BA  :  BC  : :  GF  :  EF ; 

and  therefore  (IV.  7)  GE  :  EF  : :  DE  :  EF;  wherefore  ED  is 
equal  (IV.  6)  to  FG.  Now  EF  is  common  to  the  two  trian- 
gles GEF,  DEF;  and  the  angle  GFE  is  equal  (const.)  to 
DEF ;  therefore  the  angle  EFD  is  equal  (I.  3)  to  FEG,  and 
D  to  G.  But  (const.)  the  angle  EFD  is  equal  to  ACB ;  there- 
fore ACB  is  equal  to  FEG ;  and  (hyp.)  the  angle  ABC  is  equal 
to  GEF  ;  wherefore,  also  (I.  20,  cor  5),  the  remaining  angles  A 
and  G  are  equal.     Therefore,  if  two  triangles,  etc. 

Prop.  VII. — Theor. — If  two  triangles  have  two  sides  of  the 
one  proportional  to  tioo  sides  of  the  other,  and  if  the  angles 
opposite  to  one  pair  of  the  homologous  sides  be  equal,  and 
those  opposite  to  the  other  pair  be  either  both  acute,  or  not 


BOOK   v.] 


EUCLID    AND    LEGENDRE. 


lit 


G 


acute,  the  angles  contained  by  the  proportional  sides  are 
equal. 

Let  ABC  and  EFG  be  two  triangles  which  have  tlie  sides 
CB,  CA  proportional  to  GF,  GE ;  the  angles  CBA,  GFE  equal, 
and  the  ether  angles  CAB,  GEF 
acute ;  then  the  angles  ACB,  EGF, 
contained  by  the  proportional  sides, 
are  equal. 

If  the  triangle  ABC  be  applied  to 
EFG,  so  that^  CB  will  foil  on  GF, 
and  the  vertex  C  on  the  vertex  G, 
and  make  GP  equal  to  CB ;  then, 
because  the  angle  CBA  is  equal  to 
the  angle  GFE,  AB  will  take  the 
direction  OP  parallel  to  EF  (I. 
16).     Since  OP  is   parallel  to  EF, 

GOP  is  equal  to  GEF  (I.  16,  cor.),  and  we  have  GF  :  GP  :: 
GE  :  GO  (V.  3) ;  but  by  hypothesis  GF  :  CB  : :  GE  :  CA,  and 
GP  is  equal  to  CB.  Hence,  CA  is  equal  to  GO  ;  therefore  OP, 
which  joins  the  extremities  of  GP  and  GO,  is  equal  to  AB, 
which  joins  the  extremities  of  CB  and  CA  (I.  ax.  1),  and  the 
triangles  ABC,  OPG  are  equal.  Hence,  the  angles  GOP, 
CAB  are  equal  (I.  16,  cor,  1),  but  the  angle  GOP  is  equal  to 
the  angle  GEF  (I.  16,  cor.  1);  consequently  (I.  14),  the  re- 
maining angle  of  GOP  is  equal  to  the  remaining  angle  of  EFG, 
and  the  angles  ACB,  EGF  are  equal.  In  the  same  manner  it 
can  be  shown  that  the  angles  ACB,  EGF  are  equal  when 
CAB,  GEF  are  not  acute.  Wherefore,  if  two  triangles  have, 
etc. 


Prop.  VIII. — Theor.  — Tn  a  right-angled  triangle,  if  a  per- 
pendicular be  drawn  from  the  right  angle  to  the  hypothenuse, 
the  triangles  on  each  side  of  it  are  simdar  to  the  whole  triangle, 
and  to  one  another.  ' 

Let  ABC  be  a  right-angled  triangle,  having  the  riijht  angle 
BAC;  and  from  the  point  A  let  AD  be  drawn  perpendicular  to 
the  hypothenuse  BC ;  the  triangles  ADB,  ADC  are  sitnilai-  to 
the  whole  triangle  ABC,  and  to  one  another. 

Because  the  angle  BAC  is  equal  (I.  ax.  11)  to  ADB,  each  of 


118  THE   ELEMENTS   OF  [bOOK   V. 

them  beiiiff  a  right  anejle,  and  that  the  ansjle  B  is  common  to 
the  two  triangles  ABC,  ABD ;  the  remaining  angle  C  is  equal 

(I.  20,  cor.  5)  to  the  remaining  angle 
BAD.  Therefore  the  triangles  ABC, 
ABD  are  equiangular,  and  (V.  3)  the 
sides  about  their  equal  angles  are 
proportionals ;  wherefore  (V.  def,  l) 
the  triancfles  are  similar.  In  the 
same  manner  it  mioht  be  demon- 
strated,  that  the  triangle  ADC  is  equiangular  and  similar  to 
ABC ;  and  the  triangles  ADB,  ADC,  being  each  equiangular 
to  ABC,  are  (I.  ax.  1)  equiangular,  and  therefore  (V.  3  and 
def.  1)  similar  to  each  other.     Therefore,  etc. 

Cor.  From  this  it  is  manifest,  that  the  perpendicular  drawn 
from  the  right  angle  of  a  right-angled  triangle  to  the  hypothe- 
nuse,  is  a  mean  proportional  (IV.  def  9)  between  the  segments 
of  the  hypothenuse;  and  also  that  each  of  the  sides  is  a  mean 
proportional  between  the  hypothenuse  and  its  segment  adjacent 
to  that  side.     For  (V.  3)  in  the  triangles  BDA,  ADC, 

as  BD  :  DA  : :  DA  :  DC ;  in  the  triangles  ABC,  DBA, 

as  BC  :  BA  : :  BA  :  BD  ;  and,  in  the  triangles  ABC,  ACD, 

as  BC  :  CA  : :  CA  :  CD. 

Scho.  This  proposition  affords  an  easy  way  of  solving  the 
first  corollary  to  the  twenty-fourth  proposition  of  the  first  book, 
as  follows: 

Let  ABC  be  a  triangle,  right-angled  at  A ;  the  square  of  the 
hypothenuse  BC  is  equivalent  to  the  squares  of  the  legs  AB, 
AC. 

Draw  AD  perpendicular  to  BC.  Then  (V.  8,  cor.)  BC  : 
BA  : :  BA  :  BD,  and  BC  :  CA  : :  CA  :  CD.  Hence  (IV.  2, 
cor.  1)  the  rectangle  BC.BD  is  equivalent  to  the  square  of  AB, 
and  the  rectangle  BC.CD  to  the  square  of  AC.  Hence  (I. 
ax.  2)  BC.BD  +  BC.CD,  or  (II.  2)  BC==^AB-+ACl 

Prop.  IX. — Puob. —  To  find  a  third  proportioyial  to  two 
given  straight  lines. 

Let  A  and  B  be  two  given  straight  lines ;  it  is  required  to 
find  a  third  proportional  to  them. 


EUCLID    AND    LEGENDKE. 


119 


BA 


BOOK   v.] 

Take  two  straight  lines  CF,  CG,  containing  any  angle  C ; 
and  upon  these  make  CD  equal  to  A,  and  DF,  CE  each  equal 
to  B.     Join  DE,  and  (I.  18)  draw  P'G  parallel 
to  it.     EG  is  the  third  proportional  required. 

For  (V.  2)  since  DE  is  parallel  to  FG,  CD  : 
DF  : :  CE  :  EG.  But  (const.)  CD  is  equal  to 
A,  and  DF,  CE  each  equal  to  B;  therefore  A  : 
B  ::  B  :  EG;  wherefore  to  A  and  B  the  third 
proportional  EG  is  found,  which  was  to  be 
done. 

aS'c'/^o,  Other  modes  of  solving  this  problem 
may  sometimes  be  employed  with  advantage. 
The  following  are  among  the  most  useful : 

1.  Draw  AD  (fig.  to  prop.  8)  perpendicular  to  the  indefinite 
straight  line  BC,  and  make  DB,  DA  equal  to  the  given  lines ; 
join  AB,  and  draw  AC  perpendicular  to  it ;  DC  is  the  third 
proportional  required.     For  (V.  8,  cor.)  BD  :  DA  :  :  DA  :  DC. 

2.  Draw  BC  perpendicular  to  AB,  and  having  made  BAand 
AC  equal  to  the  less  and  greater  of  the  given  lines,  draw  CD 
and  BE  perpendicular  to  AC  ;  AD  will  be  a  third  proportional 
to  AB  and  AC,  and  AE  to  AC  and  AB.  This  follows  from  the 
third  proposition  of  this  book,  since  the  triangles  ABC,  ACD 
are  equiangular,  as  are  also  ABC,  AEB. 

The  angles  ABC,  ACD,  etc.,  are  hex-e  made  right  angles. 
They  may  be  of  any  magnitude, 
however,  provided  they  be  equal. 
It  is  sufficient,  therefore,  to  draw 
two  straight  lines,  AB,  AC,  making 
any  angle  ;  to  cut  off  AB,  AC  equal 
to  the  given  ))roportionals  ;  and 
then,  BC  being  joined,  to  make  the 

angle  ACD  equal  to  ABC,  and  to  draw  BE  parallel  to  CD;  or 
to  make  the  angle  ABE  equal  to  AL'B,  and  to  draw  CD  paral- 
lel to  BE. 

This  method  affords  an  easy  means  of  continuing  a  vseries  of 
lines  in  contiimal  proportion,  both  ways,  when  any  two  succes- 
sive terms  are  given.  Thus,  after  CD  and  1)E  are  drawn,  it  is 
only  necessary  to  draw  DF,  EG,  etc.,  parallel  to  BC,  and  F'H, 
GK,  etc.,  parallel  to  CD ;  as  AD,  x\F,  iVH,  etc.,  will  be  the  siic- 


G    B    D    H 


120 


THE   ELEMENTS    OF 


[cook  V. 


A  C  B 


ceedinpj  terms  of  the  ascending  scries,  and  AE,  AG,  AK,  etc., 
those  of  the  desceiidiiify  one. 

Prop.  X. — Prob. — To  find  a  fourth  proportional  to  three 
given  straight  lines. 

Let  A,  B,  C  be  three  given  straight  lines;  it  is  required  to 
find  a  fourth  proportional  to  them. 

Take  two  straight  lines  DE,  DF,  contain- 
ing any  angle  EDF,  and  make  DG  equal  to 
(  A,  GE  equal  to  B,  and  DH  equal  to  C  ;  and 
having  joined  GH,  draw  (I.  18)  EF  parallel 
to  it  through  the  point  E;  HF  is  the  fourth 
proportional  required. 

For  (V.  2)  since  Gil  is  parallel  to  EF,  as 

DG  :  GE  : :  DH  :  HF;  but  DG  is  equal  to 

A,  GK  to  B,  and  DH  to  C ;  therefore  as  A : 

B  : :  C  :  HF;  wherefore,  to  the  three  given 

straight  lines.  A,  B,  C,  the  fourth  proportional  HF  is  found ; 

which  was  to  be  done. 

^cho.  1.  The  solution  of  this  problem  may  also  be  effected  in 
several  different  ways;  some  of  which  may  be  meiely  indicated 
to  the  student,  as  the  proofs  present  no  difficulty ;  and  it  is  evi- 
dent that,  with  slight  modification,  they  are  applicable  in  solv- 
ing the  ninth  proposition,  which  is  only  a  particular  case  of  the 
tenth. 

1.  Let  AE,EC  (last  fig.  to  HL  20)  be  the  second  and  third 
terms,  placed  contiguous,  and  in  the  same  straight  line,  and 
draw  BE,  making  any  angle  with  AC,  and  equal  to  the  first 
term  ;  through  the  three  jioints,  A,  B,  C,  describe  a  circle  cut- 
ting BE  produced  in  D;  ED  is  the  fourth  proportional.  If 
AB  and  CD  be  joined,  the  proof  will  be  obtained  by  means  of 
the  triangles  ABE,  CDE,  which  are  similar. 

2.  Draw  AB,  AC  (fig.  to  IH.  21,  cor.)  making  any  angle,  and 
make  AB,  AE  equal  to  the  second  and  third  tei'uis;  then  if  AC 
be  taken  equal  to  the  first  term,  and  a  circle  be  described  pass- 
ing thiough  B,  C,  and  E,  and  meeting  AC  in  F,  AF  will  be 
the  required  line.  If  BF  and  EC  be  joined,  the  triangles  ABF 
and  ACE  are  similar,  and  hence  the  proof  is  immediately  ob- 
tained. 


BOOK  v.]  EUCLID  AND  LEGENDE"S.  121 

3,  Make  BD  and  DA  (fig.  to  V.  8)  perpendicular  to  each 
otlicT,  and  equal  to  tlie  first  and  second  terms;  join  AB,  and 
draw  AC  perpendicular  to  it;  in  DA,  produced  through  A,  if 
necessary,  take  a  line  equal  to  the  third  term,  through  the 
upper  extremity  of  which  draw  a  line  parallel  to  AC  ;  the  line 
intercepted  on  DC,  produced  if  necessary,  hetween  D  and  tliis 
parallel,  is  the  fourth  proportional  required.  ^ 

Cor.  1.  If  four  straight  lines  be  proportionals,  the  rectangle 
contained  by  the  extremes  is  equivalent  to  that  conta,ined  by 
the  means;  and  (2)  if  the  rectangle  contained  by  the  extremes 
be  equivalent  to  that  contained  by  the  means,  the  four  straight 
lines  are  proportionals  (V.  2,  cor.). 

iScko.  2.  This  corollary,  of  which  the  corollary  immediately 
following  is  a  case,  affords  the  means  of  deriving  the  equality 
of  rectangles,  and  the  proportion  of  straight  lines  containing 
them,  from  one  another.  It  evidently  corresponds  to  IV. 
prop.  1,  cor.,  and  prop.  2,  cor.  1,  and  it  might  be  regarded  as  an 
immediate  result  of  those  corollaries,  without  any  distinct 
proof,  if  the  lines  were  expressed  (T-.  2-3,  cor.  4)  by  lineal  units, 
and  the  rectangles  by  superficial  ones.  This  corollary  and  the 
third  proposition  of  this  book,  when  employed  in  connection 
with  one  another,  form  one  of  the  most  powerful  instruments 
in  geometrical  investigations,  and  they  facilitate  in  a  peculiar 
degree  .the  application  of  algebra  to  such  inquiries. 

Cor.  2.  If  three  straight  lines  be  proportionals,  the  rectangle 
contained  by  the  extremes  is  equivalent  to  the  square  of  the 
mean  ;  and  (2)  if  the  rectangle  contained  by  the  extremes  be 
equivalent  to  the  square  of  the  mean,  the  three  straight  lines 
are  proportionals. 

Prop.  XI. — Prob.  —  To  find  a  mean  proportional  hetween  two 
given  straight  lines. 

Let  AB,  BC  be  two  given  straight  lines ;  it  is  required  to  find 
a  mean  proportional  between  them. 

Place  AB,  BC  in  a  straight  line,  and  upon  AC  as  diameter 
describe  the  semicircle  ADC ;  fi-om  B  (T.  7)  draw  BD  at  right 
angles  to  AC  ;  BD  is  the  mean  proportional  between  AB  and 
BC. 

Join  AD,  DC.     Then,  because  the  angle  ADC  in  a  semicircle 


122  THE    ELEMENTS   OF  [bOOK   V. 

is  (III.  11)  a  right  angle,  and  because  in  the  right-angled  tri- 
angle ADC,  DB  is  drawn  from  the  right 
angle  perpendicular  to  AC,  DB  is  a  mean 
proportional  (V.  8,  cor.)  between  AB,  BC, 
the  segments  of  the  base.  Therefore  be- 
tween AB,  BC,  the  mean  proportional  DB 
is  found  ;  Avhich  was  to  be  done. 
Scho.  Out  of  several  additional  ways  of  solving  this  prob- 
lem, the  following  may  be  mentioned : 

1.  On  the  greater  extreme  as  diameter  describe  a  semicircle  ; 
from  the  diameter  cut  oif  a  segment  equal  to  the  less  extreme, 
through  the  extremity  of  which  draw  a  perpendicular  cutting 
the  circumference ;  and  the  chord  drawn  from  that  intersection 
to  the  extremity  of  the  diameter  common  to  the  two  extremes 
is  the  required  mean.  The  proof  is  manifest  from  III.  11,  and 
V.  8,  cor. 

2.  Make  AD,  DC  (2d  fig.  to  III.  21)  equal  to  the  given  ex- 
tremes ;  on  AC  as  chord  describe  any  circle,  and  a  tangent 
drawn  from  D  will  be  the  required  mean.  The  proof  of  this  is 
obtained  by  joining  AB  and  BC,  as  the  triangles  ADB,  BDC 
are  similar. 

When  one  mean  is  determined,  others  may  be  found  between 
it  and  the  given  extremes,  and  thus  three  means  will  be  insert- 
ed between  the  given  lines ;  and  by  finding  means  between 
each  successive  pair  of  the  five  terms  of  which  the  series  then 
consists,  the  number  of  means  will  be  increased  to  seven.  By 
continuing  the  process  Ave  may  find  fifteen  nieans,  thirty-one 
means,  or  any  number  which  is  less  by  one  than  a  power  of  2. 
We  cannot  find,  however,  by  elementary  geometry,  any  other 
number  of  means,  such  as  two,  four,  or  five. 

Cor.  A  given  straight  line  can  be  divided  in  extreme  and 
mean  ratio. 

Prop.  XTI. — Theor. — J^quivalent  pnrnllelograms  inhich  Imve 
an  an.fjle  of  the  one  equal  to  an  angle  of  the  other,  have  their 
sides  about  those  angles  reciprocally  i/roportioiHil ;  atid  (2) 
parallelograms  lohich  have  an  angle  of  the  one  equal  to  an  an^ 
gle  of  the  <  titer,  and  the  sides  a,bout  those  angles  reciprocally 
proportional.,  are  equioale^it  to  one  another. 


BOOK   v.]  EUCLID   AND    LEGENDKE.  123 

1.  Let  AB,  BC  be  equivalent  parallelograms,  which  have 
the  angles  at  B  equal ;  the  sides  about  those  angles  are  recip- 
rocally proportional;  that  is,  DB  :  BE  : :  GB  :  BF. 

Let  the  sides  DB,  BE  be  placed  in  the  same  straight  line, 
and  contiTUOus,  and  let  the  parallelograms  be  on  opposite  sides 
of  DE ;  then  (I.  10,  cor.)  because  the  angles  at  B  are  equal, 
FB,  BG  are  in  one  straight  line.  Complete  the  parallelogram 
FE,  and  (IV.  5)  because  AB  is  equal  to  BC,  and  that  FE  is 
another  parallelogram, 

AB:  FE::BC:FE.     But  (V.  1) 
as  AB  to  FE,  so  is  the  base  DB  to  BE  ;  and 
as  BC  to  FE,  so  is  the  base  GB  to  BF ; 
therefore  (IV.  1)  as  DB  :  BE  : :  GB  :  BF.     The  sides,  therefore, 
of  the  parallelograms  AB,  BC,  about  their 
equal  angles,  are  (V.  def   2)  reciprocally 
proportional. 


2.  But  let  the  sides  about  the  equal  an- 
gles be  reciprocally  proportional,  viz.,  DB  : 
BE  : :  GB  :  BF ;  the  parallelograms  AB, 
BC  are  equivalent. 

The  same  construction  being  made,  because, 
as  DB  :  BE  :  :  GB  :  BF;  and  (V.  1) 
as  DB  :  BE  : :  AB  :  FE ;  and  as  GB  :  BF  : :  BC  :  FE ; 
therefore  (IV.  1)  as  AB  :  FE  : :  BC  :  FE  ;  wherefore  (IV.  6) 
the  parallelogram  AB  is  equivalent  to  the  parallelogram  BC. 
Therefore  equivalent  parallelograms,  etc. 

Scho.  1.  If  AD,  CG  were  produced  to  meet,  it  would  be  easy 
to  show,  that  AB  and  BC  would  be  the  complements  of  the 
parallelograms  about  the  diagonal  of  the  whole  parallelogram 
AC. 

In  the  demonstration,  it  should  in  strictness  be  proved  that 
AF  and  CE  meet  when  produced.     This  follows  from  I.  19. 

Cor.  Hence,  equivalent  triangles  which  have  an  angle  of  the 
one  equal  to  an  angle  of  the  other,  have  their  sides  about  those 
angles  reciprocally  proportional ;  and  (2)  triangles  which  have 
an  angle  of  the  one  equal  to  an  angle  of  the  other,  and  the  sides 
about  those  angles  reciprocally  proportional,  are  equivalent  to 
one  another. 

JScho.  2.  It  is  evident  from  the  seventh  corollary  to  the  fif- 


124 


THE    ELEMENTS   OF 


[book  V. 


teenth  proposition  of  the  first  book,  that  this  proposition  is 
true  as  well  as  when  the  angles  are  supplemental  as  when  they 
are  equal. 


Prop.  XIIT. — Prob. —  TIpon  a  given  straight  line  to  describe 
a  figure  similar  to  a.  given  rectilineal  fi  ure,  and  such  that  the 
given  line  shall  be  homologous  to  an  assigned  side  of  the  given 
figure. 

Let  AB  be  the  given  straight  line,  on  which  it  is  reqnirefi  to 
describe  a  rectilineal  figure  similar  to  a  given  rectilineal  figure, 
and  such  that  AB  may  be  homologous  to  CD,  a  side  of  the 
given  figure. 

First,  let  the  given  rectilineal  figure  be  the  triangle  CDE. 
Make  the  angles  BAF,  ABF  respectively  equal  to  DCP],  CDE; 
and  (V,  3,  cor.)  the  triangle  ABF  is  similar  to  CDE,  and  has 
AB  homologous  to  CD. 

Again:  let  the  given  figure  be  the  quadrilateral  CDOE. 
Join  DE,  and,  as  in  the  first  part,  describe  the  triangle  ABF 
having  the  angles  BAF,  ABF  respectively  equal  to  DCE,  CDE, 

and  also  the  triangle  BFH  having  the 
angles  FBH,  BFH  respectively  equal 
to^EDG,  DEC.  Then  (I.  20,  cor.  5) 
the  angles  BHF,  DGE  are  equal,  and 
(const.)  A  and  C  are  equal.  Also,  since 
(conet.)  ABF,  FBH  are  respectively 
equal  to  CDE,  EDG,  the  whole  ABH 
is  equal  to  the  whole  CDG.  For  the 
same  reason  AFH  is  equal  to  CEG  ;  and  therefore  the  quadri- 
lateral figui-es  A  BHF,  CDGE  are  equiangular.  But  likewise 
these  figures  have  their  sides  about  the  equal  angles  propor- 
tional. For  the  triangles  ABF,  CDE  being  equiangular,  and 
also  BFH,  DEG;  as  BA  :  AF  :  :  DC  :  CE;  and^as  FH.  : 
nB::EG:  GD.  Also,  in  the  same  triangles,  AF  :  FB  :: 
CE  :  ED  ;  and  as  FB  :  FH  : :  ED  :  EG  ;  therefore,  ex  mquo, 
AF  :  FH  :  :  CE  :  EG.  In  the  same  maimer  it  may  be  pi'oved, 
that  AB  :  BH  :  :  CD  :  DG;  whcrefoi-e  (V.  def  1)  the  figures 
ABIIF,  CDGE  are  similar  to  one  another. 

Next :  let  the  given  figure  be  CDKGE.  Join  DG  ;  and,  as  in 
the  second  case,  deiscribe  the  figure  ABHF,  similar  to  CDGE, 


BOOK  v.]  EUCLID  AND  LKGENDRE.  125 

and  similarly  situated;  also,  describe  the  triangle  BHL  having 
the  angles  BHL,  IIBL  respectively  equal  to  DGK,  GDK.  Then 
(I.  ax.  2)  the  whole  angles  FIIL,  ABL  are  respectively  equal 
to  the  whole  angles  EGK,  CDK ;  and  (I.  20,  cor.  5)  the  angles 
L  and  K  are  equal.  Therefore  the  figures  ABLHF,  CUKGE 
are  equiangular.  Again :  because  the  quadrilaterals  ABHF, 
CDGE,  and  the  triangles  BLH,  DKG  are  similar;  as  FH  : 
HB  ::  EG  :  GD,  and  HB  :  HL  : :  GD  :  GK  ;  therefore,  ex 
cequo,  FH  :  HL  :  :  EG  :  GK.  In  like  manner  it  may  be  shown, 
that  AB  :  BL  : :  CD  :  DK  ;  and  because  the  quadrilateials 
ABHF,  CDGE,  and  the  triangles  BLH,  DKG  are  similar,  the 
sides  ?,bout  the  angles  A  and  C,  L  and  K,  AFH  and  CEG  are 
proportional.  Therefore  (V.  def  l)  the  five-sided  figures 
ABLHF,  CDKGE  are  similar,  and  the  sides  AB  and  CD  are 
homologous.  In  the  same  manner,  a  rectilineal  figure  of  six  or 
more  sides  may  be  described  on  a  given  straight  line,  similar  to 
one  given  ;  which  was  to  be  done. 

Scho.  In  practice,  if  AB  be  parallel  to  CD,  the  cojistruction 
is  most  easily  efll-cted  by  drawing  AF  and  BF  parallel  to  CE 
and  DE  ;  then  FH  and  BH  parallel  to  EG  and  DG  ;  and  lastly, 
HL  and  BL  parallel  to  GK  and  DK.  The  doing  of  this  is 
much  facilitated  by  employing  tlie  useful  instrument,  the  par- 
allel ruler.  For  the  easiest  methods,  however,  of  performing 
this  and  many  other  problems,  the  student  must  have  recourse 
to  works  that  treat  expressly  on  such  subjects,  particulai'ly 
treatises  on  practical  geometry,  surveying,  and  the  use  of 
mathematical  instruments. 

Prop.  XIV. — Theor. — Similar  plane  figures  are  to  one  an- 
other in  the  duplicate  ratio  of  their  homologous  sides. 

Let  ABC,  DEF  be  sitnilar  triangles,  having  the  angles  B  and 
E  equal,  and  AB  :  BC  ::  DE  :  EF,  so  that  (IV.  def.  13)  the 
side   BC    is    homologous    to   EF.  A 

The  ti'iangle  ABC  has  to  the  trian- 
gle  DEF  the  duplicate  ratio  of 
that  which  BC  has  to  EF. 

Take  (V.  9)  BG  a  third  propor- 
tional to  BC,  EF,  so  that  BC  :  EF 
: :  EF  :  BG,  and  join  GA.     Then,  because 


126  THE   ELEITENTS   OF  [bOOK   V. 

AB  :  BC  : :  DE  :  EF ;  alternately, 
AB  :  DE  : :  BC  :  EF ;  but  (const.) 
as  BC  :  EF  : :  EF  :  BG  ; 
therefore  (IV.  V)  as  AB  :  DE  : :  EF  :  BG.  The  sides,  therefore, 
of  the  triangles  ABG,  DEF  -which  are  about  the  equal  angles, 
are  reciprocally  proportional,  and  therefore  (V.  12,  cor.)  the 
triangles  ABG,  DEF'  are  equal.  Again :  because  BC  :  EF  : : 
EF  :  BG  ;  and  that  if  three  straight  lines  be  proportionals,  the 
first  is  said  (IV,  def  11)  to  have  to  the  third  the  duplicate  ratio 
of  that  which  it  has  to  the  second  ;  BC  therefore  has  to  BG  the 
duplicate  ratio  of  that  which  BC  has  to  EF.  But  (V.  1)  asBC 
to  BG,  so  is  the  triangle  ABC  to  ABG.  Therefore  (IV.  7)  the 
triangle  ABC  has  to  ABG  the  duplicate  ratio  of  that  which  BC 
has  to  EF.  But  the  triangle  ABG  is  equal  to  DEF ;  wherefore, 
also,  the  triangle  ABC  has  to  DEF  the  duplicate  ratio  of  that 
which  BC  has  to  EF.     Therefore,  etc. 

Scho.  1.  From  this  it  is  manifest,  that  if  three  straight  lines 
be  proportionals,  as  the  first  is  to  the  third,  so  is  any  triangle 
upon  the  first  to  a  similar  triangle  similarly  described  on  the 
second. 

The  third  proportional  might  be  taken  to  EF  and  BC,  and 
placed  from  E  along  EF  produced ;  and  then  a  triangle  equal 
to  ABC  would  be  formed  by  joining  D  with  the  extremity  of 
the  produced  line. 

Scho.  2.  The  above  case  might  also  be  proved  by  making 
BH  equal  to  EF,  joining  AH,  and  drawing  through  H  a  par- 
allel to  AC.  The  triangle  cut  off  by  the  parallel  is  equal  (1. 14) 
to  DEF.  But  (V.  1)  the  triangle  ABC  is  to  the  triangle  ABH 
as  BC  to  BH;  and,  for  the  same  reason,  the  triangle  ABH  is  to 
the  triangle  cut  off  by  the  parallel,  or  to  DEF,  as  BC  to  BH,  or 
(const.)  as  BH  to  BG.  Therefore,  ex  cequo^  the  triangle  ABC 
Las  to  the  triangle  DEF  the  same  ratio  that  BC  has  to  BG,  or 
(IV.  def.  11)  the  duplicate  ratio  of  that  which  BC  has  to  EF. 

Again  :  let  ABCD  and  AEFG  be  two  squares,  then  will 
they  be  to  one  another  in  the  duplicate  ratio  of  their  sides — as 
it  has  been  previously  demonstrated  that  similar  triangles  are 
to  one  another  in  the  duplicate  ratio  of  their  liomologous  sides. 
Therefore  : 

ABD  :  AEF  in  the  duplicate  ratio  of  AB  to  AE ;  and  ADC  : 


BOOK  v.] 


EUCLID  AND  LEGENDEE. 


127 


D 

/ 

/ 

F 


AFG  in  the  duplicate  ratio  of  AB  to  AE ;  hence,  by  compo- 

pition,  ABD+ADC  :  AEF+ AFG  in  the 

duplicate  ratio  of  AB  to  AE  ;  but  ABD 

+  ADC=ABDC,    and    AEF+AFG  = 

AEFG ;    wherefore  the  squares  are   to 

one  another  in  the  duplicate   ratio  of 

their  sides. 

For  like  reason,  the,  triangles  AIID, 
AFC  are  to  one  another  in  the  duplicate 

ratio  that  PH  is  to  DF,  or  that  AD  is  to  AC ;  hence,  similar 
triangles  are  one  to  another  in  the  duplicate  ratio  of  their  alti- 
tudes or  bases ;  since  the  segments  AHD,  AFC  have  the  same 
bases  and  altitudes  as  the  triangles  AHD  and  AFC,  and  being 
segments  of  quadrants,  are  similar ; 
hence,  they  are  to  one  another  in  the  du- 
plicate ratio  of  their  bases  and  altitudes ; 
therefore,  by  composition,  AED  +  seg. 
AHD :  ABC  +  seg.  AFC  in  the  duplicate 
ratio  of  their  homologous  sides,  or  simi- 
lar polygons  are  to  one  another  in  the 
duplicate  ratio  of  their  homologous  sides, 
and  quadrants  of  circles  are  one  to  an- 
other in  the  duplicate  ratio  of  their  radii ;  hence,  semicircles 
are  one  to  another  in  the  duplicate  ratio  of  their  diameters,  and. 
circles  are  one  to  another  in  the  duplicate  ratio  of  their  diame- 
ters.    Wherefore,  similar  surfaces  are,  etc. 

Cor.  1.  Therefore,  universally,  if  three  lines  be  proportionals, 
the  first  is  to  the  third  as  any  plane  figure  upon  the  first  to  a 
similar  and  similarly  described  figure  upon  the  second. 

Cor.  2.  Because  all  squares  are  similar  figures,  and  all  circles 
are  similar  figures,  the  ratio  of  any  two  squares  to  one  another 
is  the  same  as  the  duplicate  ratio  of  their  sides ;  and  the  ratio 
of  any  two  circles  to  one  another  is  the  same  as  the  duplicate 
ratio  of  their  diameters ;  hence,  any  two  similar  plane  figures 
are  to  one  another  as  the  squares  or  circles  (IV.  7)  described  on 
their  homolocrous  sides. 

Cor.  3.  Because  the  sides  of  similar  plane  figures  are  propor- 
tional, therefore  (IV.  8)  their  perimeters  or  peripheries  are 
propoi'tional  to  the  homologous  sides ;  hence,  the  perimeters  of 


128  THE   ELKMENTS    OF  [bOOK   V. 

similar  polyo^ons  are  proportional  to  their  apotliems ;  the  cir- 
cumfei-enees  of  circles  are  proportional  to  their  diameters. 

Cor.  4.  Hence  plane  figures  which  are  similar,  to  the  same 
figure  are  similar  to  one  another  (I.  ax.  l). 

Cor.  5.  If  four  straight  lines  be  proportionals,  the  similar 
plane  figures,  similarly  desci-ibed  upon  them,  are  also  propor- 
tionals; and,  conversely.,  if  the  similar  plane  figures,  similarly- 
described  upon  four  straight  lines,  be  proportionals,  those  lines 
are  proportionals. 

Cor.  6.  Similar  polygons  inscribed  in  circles  are  to  one  an- 
other as  the  squares  of  the  diameters. 

Cor.  1.  When  there  are  three  parallelograms,  AC,  CH,  CF, 
the  first,  AC  (IV.  def  10),  has  to  the  third,  CF,  the  ratio  which 
is  compounded  of  the  ratio  of  the  first,  AC,  to  the  second,  CIT, 
and  of  the  ratio  of  CH  to  the  third,  CF;  but  AC  is  to  CH  as 
their  bases  ;  and  CH  is  to  CF  as  their  bases  ;  therefore  AC  has 
to  CF  the  ratio  which  is  compounded  of  ratios  that  are  the 
same  with  the  ratios  of  the  sides. 

Scho.  3.  Dr.  Simson  remarks  in  his  Note  on  this  corollaiy, 
that  "nothing  is  usually  reckoned  more  difficult  in  the  elements 
of  geometry  by  learners,  than  the  doctrine  of  compound  ratio." 
This  distinguished  geometer,  however,  has  both  freed  the  text 
of  Euclid  from  the  errors  introduced  by  Theon  or  others,  and 
has  explained  the  subject  in  such  a  manner  as  to  remove  the 
difficulties  that  were  formeily  felt.  According  to  liim,  "  every 
proposition  in  which  compound  ratio  is  made  use  of,  may  with- 
out it  be  both  enunciated  and  demonstrated  ;"  and  "  the  use  of 
compound  ratio  consists  wholly  in  this,  that  by  means  of  it,  cir- 
cumlocutions may  be  avoided,  and  thereby  propositions  may- 
be more  briefly  either  enunciated  or  demonstrated,  or  both 
may  be  done.  For  instance,  if  this  corollary  were  to  be  enun- 
ciated, without  mentioning  compound  ratio,  it  might  be  done 
as  follows:  If  two  pai-allelograms  be  equiangular,  and  if  as  a 
side  of  the  first  to  a  side  of  the  second,  so  any  assumed  straight 
line  be  made  to  a  second  straight  line ;  and  as  the  other  side 
of  the  first  to  the  other  side  of  the  second,  so  the  second 
straight  line  be  made  to  a  third  ;  the  first  parallelogram  is  to 
the  second  as  the  first  straight  line  to  the  third  ;  and  the 
demonstration  would  be  exactly  the  same  as  we  now  have  it. 


BOOK  v.]  EUCLTD  AND  LEGENDKE.  129 

But  the  ancient  cceometers,  when  they  observed  this  ennncia- 
ti(ni  could  l)e  made  nhortev,  l)y  giving  a  name  to  the  ratio 
which  the  first  straiglit  line  lias  to  the  last,  by  which  name  the 
intermediate  ratios  miglA  likewise  be  signified,  of  the  first  to 
the  second,  and  of  the  second  to  the  third,  and  so  on,  if  there 
■were  more  of  them,  they  called  this  ratio  of  the  fiist  to  the  last, 
the  ratio  compounded  of  the  ratios  of  the  first  to  the  second, 
and  of  the  second  to  the  third  straight  line;  that  is,  in  the 
present  example,  of  the  ratios  which  are  the  same  with  the 
ratios  of  the  sides." 

Scho.  4.  The  seventh  corollary  will  be  illustrated  by  the  fol- 
lowing proposition,  which  exhibits  the  subject  in  a  different, 
and,  in  some  respects,  a  preferable  light : 

Triangles  which  have  an  angle  of  the  one  equal  to  an  angle 
of  the  othei\  are  proportional  to  the  rectangles  covitcined  by 
the  sides  about  those  angles ;  and  (2)  equiangular  parallelO' 
grams  are  proportional  to  the  rectangles  co7itained  by  their  ad- 
jacent  sides. 

1.  Let  ABC,  DBE  be  two  triangles,  having  the  angles  ABC, 
DBE  equal;  the  first  triangle  is  to  the  second  as  AB.BC  is  to 
DB.BE. 

Let  the  triangles  be  placed  with  their  equal  angles  coinciding, 
and  join  CD.  Then  (V.  1)  AB  is  to  DB  as  the  triangle  ABC 
to  DBC.  But  (V.  1,  cor.  2)  AB  :  DB  ::  AB.BC  :DB.BC; 
theiefore  (IV.  7)  AB.BC  is  to  DB.BC  as  the  triangle  ABC  to 
DBC.  Li  the  same  maimer  it  would 
be  shown  that  DB.BC  is  to  DB.BE  as 
the  triangle  DBC  to  DBE ;  and,  there- 
fore, ex  ceq^io^  AB.BC  is  to  DB.BE  as 
the  triangle  ABC  to  DBE. 

2.  If  parallels  to  BC  through  A  and 
D,  and  to  AB  through  C  and  E  were 

drawn,  parallelograms  would  be  formed  which  would  be  re- 
spectively double  of  the  triangles  ABC  and  DBE,  and  which 
(IV.  9)  would  have  the  same  ratio  as  the  triangles;  that  is,  the 
ratio  of  AB.BC  to  DB.BE;  and  this  proves  the  second  part  of 
the  proposition. 

Comparing  this  proposition  and  the  corollary,  we  see  that  the 
ratio  which  is  compounded  of  ihe  ratio  of  the  sides,  is  the  same 
9 


130 


THE    ELEMENTS   OF 


[book  V. 


as  the  ratio  of  their  rectangles,  or  the  same  (T.  23,  cor.  5)  as  the 
ratio  of  their  products,  if  they  he  e:q)ressed  in  numbers.  This 
conclusion  might  also  be  derived  from  the  proof  given  in  the 
text.  For  (const.)  DC  :  CE  : :  CG  :  K ;  whence  (V.  10,  cor.) 
K.DC  =  CE.CG.  But  it  was  proved  that  BC  :  K  : :  AC  :  CF; 
or  (V.  1)  BC.DC  :  K.DC  : :  AC  :  CF;  or  BC.DC  :  CE.CG  :: 
AC  :  CF;  because  K.DC=CE.CG. 

The  twelfth  proposition  of  this  book  is  evidently  a  case  of 
this  proposition;  and  the  fourteenth  is  also  easily  derived 
from  it. 


Pijop.  XV. — TriEOPw — The  parallelograms  ahout  the  diago- 
nal of  any  parallelogram  are  similar  to  the  whole,  and  to  one 
another. 

Let  ABCD  be  a  parallelogram,  and  EG,  HK  the  parallelo- 
grams about  the  diagonal  AC  ;  the  parallelograms  EG,  HK  are 
similar  to  the  whole  parallelogram,  and  to  one  another. 

Because  DC,  GF  are  parallels,  the  angles  ADC,  AGF  are 

(I.  16)  equal.  For  the  same  reason,  be- 
cause BC,  EF  are  parallels,  the  angles 
ABC,  AEF  are  equal ;  and  (T.  15,  cor.  1) 
each  of  the  angles  BCD,  EFG  is  equal 
to  the  opposite  angle  DAB,  and  there- 
fore they  are  equal  to  one  another ; 
Avherefore,  in  the  parallelograms,  the 
angle  ABC  is  equal  to  AEF,  and  BAC  common  to  the  two  tri- 
angles BAC,  EAF ;  therefore  (V.  3,  cor.)  as  AB  ;  BC  : :  AE  : 
EF.  And  (IV.  5),  because  the  opposite  sides  of  parallelograms 
are  equal  to  one  another,  AB  :  AD  ::  AE  :  AG;  and  DC  : 
BC  : :  GF  :  EF;  and  also  CD  :  DA  : :  FG  :  GA.  Therefore 
the  sides  of  the  parallelograms  BD,  EG  about  the  equal  angles 
are  proportionals;  the  parallelograms  are,  therefore  (V.  def  l), 
similar  to  one  another.  In  the  same  manner  it  would  be  shown 
that  the  parallelogram  BD  is  similar  to  HK.  Therefore  each 
of  the  parallelograms  EG,  HK  is  similar  to  BD.  But  (V.  14, 
cor.)  rectilineal  figures  which  are  similar  to  the  same  figure,  are 
similar  to  one  another;  therefore  the  parallelogram  EG  is  simi- 
lar to  HK.     Wherefore,  etc. 

Scho.  Hence,  GF  :  FE  : :  FH  :  FK.    Therefore  the  sides  of 


BOOK  v.] 


EUCLID  AND  LEGENDRE. 


131 


the  paralleloojrains  GK  and  EH,  about  the  equal  angles  at  F, 
are  reciprocally  proportional ;  and  (V.  12)  these  parallelograms 
are  equivalent ;  a  conclusion  which  agrees  with  the  eighth 
corollary  to  the  fifteenth  proposition  of  the  first  book. 

Prop.  XVI. — Prob. —  To  describe  a  rectllmeal  figure  which 
shall  be  similar  to  one  given  rectilineal  figure.,  and  equivalent 
to  one  another. 

Let  ABC  and  D  be  given  rectilineal  figures.  It  is  required 
to  describe  a  figure  similar  to  ABC,  and  equal  to  D. 

Upon  the  straight  line  BC  describe  (II.  5,  scho.)  the  parallelo- 
gram BE  equivalent  to  ABC  ;  also  upon  CE  describe  the  parallel- 
ogram CM  equivalent  to  D,  having  the  angle  FCE  equal  to 
CBL.  Therefore  (I.  16  and  10)  BC  and  CF  are  in  a  straight 
line,  as  also  LE  and  EM.  Between  BC  and  CF  find  (V.  11)  a 
mean  proportional  GH,  and  ^ 

on  it  describe  (V.  13)  the 
figure  GHK  similar,  and 
similarly  situated,  to  ABC; 
GHK  is  the  figure  required. 

Because  BC  :  GH  : :  GH  : 
CF,  and  if  three  straight 
lines  be  proportionals,  as  the 
first  is  to  the  third,  so  is  (V. 

14,  cor.  2)  the  figure  upon  the  first  to  the  similar  and  similarly 
described  figure  upon  the  second  ;  therefore, 

as  BC  to  CF,  so  is  ABC  to  KGH  ;  but  (V.  1) 
as  BC  to  CF,  so  is  BE  to  EF  ; 
therefore  (IV.  7)  as  ABC  is  to  KGH,  so  is  BE  to  EF.  But 
(const.)  ABC  is  equivalent  to  BE  ;  therefore  KGH  is  equivalent 
(IV.  ax.  4)  to  EF;  and  (const.)  EF  is  equivalent  to  D;  where- 
fore, also,  KGH  is  equivalent  to  D;  and  it  is  similar  to  ABC. 
Therefore  the  rectilineal  figure  KGH  has  been  described,  simi- 
lar to  ABC  and  equivalent  to  D  ;  which  was  to  be  done. 


Prop.  XVII. — Theor. — If  two  similar  parallelograms  have 
a  comm,on  angle,  and  be  similarly  situated,  they  are  about  the 
same  diagonal. 

Let  ABCD,  AGEF  be  two  similar  parallelograms  having  a 


132 


THE   ELEMENTS   OF 


[book  V. 


C 


E 


F 


B 


common  angle  CAB,  tlicy  will  be  about   the   same  diagonal 

AD.  Similar  parallelograms  have 
their  sides  about  equal  angles  pro- 
portional (V.  def.  l).  Draw  the 
diagonals  EG  and  CB  ;  hence,  AB  : 
AG  : :  AC  :  AE  ;  therefore  EG  ia 
parallel  to  CB  (V.  3),  and  the  angles 
AEG,  ACB  are  equal  (I.  16);  like- 
wise the  angles  EGA,  CBA.  The 
triangles  E AG,  FG A  are  equal  (I.  15,  cor.  5);  likewise  the  tri- 
angles CAB,  DBA  ;  therefore  AF  is  equal  to  EG,  and  AD  is 
equal  to  CB  ;  but  AF  is  the  diagonal  also  of  AGEF,  and  is  in 
the  same  straight  line  with  AD,  the  diagonal  of  ABCD. 
Wherefore,  if  two  similar  parallelograms,  etc. 

Cor.  Hence,  equiangular  parallelograms  have  to  one  another 
the  ratio  which  is  compounded  of  the  ratio  of  their  sides; 
hence,  triangles  which  have  one  angle  of  the  one  equal,  or  sup- 
plemental, to  one  angle  of  the  other,  have  to  one  another  the 
ratio  which  is  compounded  of  the  ratio  of  the  sides  containing 
those  angles. 


Pkop.  XVIII. — ^Theok. —  Of  all  the  parallelof/rmns  that  can 
he  inscribed  in  any  triangle^  that  which  is  described  on  the  half 
of  one  of  the  sides  as  base  is  the  greatest. 

Let  ABC  be  a  triangle,  having  BC,  one  of  its  sides,  bisected 
in  D;  draw  (I.  18)  DE  parallel  to  BA,  and  EF  to  BC  ;  let  also 

G  be  any  other  point  in  BC,  and 
describe  the  parallelogram  GK; 
FD  is  greater  than  KG. 

If  G  be  in  DC,  through  C  draw 
CL  parallel  to  BA,  and  produce 
FE,  KH,  Gil  as  in  the  figure. 
Then  (I.  15,  cor.  8)  the  comj>le- 
ments  LTI  and  IID  are  equivalent ;  and  since  the  bases  CD, 
DB  are  equal,  the  parallelograms  ND,  DK  (I.  15,  cor.  5)  are 
equivalent.  To  LH  add  ND,  and  to  HD  add  DK  ;  then  (I. 
ax.  2)  the  gnomort  MND  is  equivalent  to  the  parallelogram 
KG.     But  (L  ax.  9)  DL  is  greater  than  MXD ;  and  therefore 


BOOK  V.J  EUCLID  AND  LFGENDEE.  133 

FD,  which  (T.  15,  cor.  5)  is  equal  to  DL,  is  greater  than  KG, 
which  is  equivalent  to  the  gnomon  JNIND.  - 

If  G  were  in  BD,  since  BD  is  equal  to  DC,  AE  is  equal  (V. 
2)  to  EC,  and  AF  to  FB;  and  by  drawing  through  A  a  jiaral- 
lel  to  BC,  meeting  DE  produced,  it  would  he  proved  in  the 
same  manner  that  FD  is  greater  than  the  inscribed  ])aralk'lo- 
gram  applied  to  BG.     Therefore,  of  all  the  parallelogi-ams,  etc. 

Cor.  Since  (V.  lo)  all  parallelograms  having  one  angle  coin- 
ciding with  BCL,  and  their  diagonals  with  CA,  arc  similar,  it 
follows  from  this  proposition  that  if,  on  the  segments  of  a  given 
straight  line,  BC,  two  parallelograms  of  the  same  altitude  be 
described,  one  of  them,  DL,  similar  to  a  given  parallelogram, 
the  other,  DF,  will  be  the  greatest  possible  when  the  segments 
of  the  line  are  equal. 

Scho.  The  parallelogram  FD  exceeds  KG  by  the  parallelo- 
gram OM  similar  to  DL  or  DF,  and  described  on  OH,  which 
is  equal  to  DG,  the  difference  of  the  bases  BD  and  BG.  Hence 
we  can  describe  parallelograms  equivalent  and  similar  to  given 
rectilineal  figures. 

The  enunciation  of  this  proposition  here  given  is  much  more 
simple  and  intelligible  than  that  of  Euclid,  and  the  proof  is 
considerably  shortened,  Euclid's  enunciation,  as  given  by  Dr. 
Simson,  is  as  follows:  "Of  all  pai'allelograms  applied  to  the 
same  straight  line,  and  deficient  by  parallelograms,  similar  and 
similarly  situated  to  that  which  is  described  upon  the  half  of 
the  line ;  that  which  is  applied  to  the  half,  and  is  similar  to  its 
defect,  is  the  greatest."  It  may  be  remaiked,  that  this  piopo- 
sition,  in  its  simplest  case,  is  the  same  as  the  second  corollary 
to  the  fifth  proposition  of  the  second  book. 

Prop.  XIX. — ffHEOR. — In  equal  circles^  or  in  the  same  cir- 
cle^ angles,  whether  at  the  centers  or  circumferences^  have  the 
same  ratio  as  the  arcs  on  which  they  stand  have  to  one  another^ 
so  also  have  the  sectors. 

Let,  in  the  equal  circles  ALB,  DNE,  the  angles  LGK,  KGC, 
CGB,  DHN,  NHM,  and  IMITF  be  at  the  centers,  and  the  angles 
BAG  and  EDF  be  at  the  circumferences,  then  will  those  angles 
have  to  each  other  the  same  ratio  as  the  arcs  KL,  KC,  CB,  DN, 
NM,  MF,  and  FE  have  to  one  another. 


134 


THE   ELEMENTS   OF 


[book  V. 


Since  (T.  def.  19)  all  angles  at  the  center  of  a  circle  are 
measured  by  the  arcs  intercepted  by  the  sides  of  the  angles, 

and  all  ancjles  at  the  circum- 

ference  are  subtended  by  the 

arcs  intercepted  by  the  sides 

of  the   angles,  and  (III.  16) 

equal  angles  will  have  equal 

arcs  whether  they  be  at  the 

center  or  the  circumference; 

hence,  the  same  ratio  which 

the  arcs  have  to  each  other,  will  the  angles  also  have  to  one 

another — that  is,  when  the  arcs  be  greater,  the  angles  will  be 

greater;  less,  less  ;  and  equal,  equal. 

And  since  the  sectors  are  contained  (III.  def  V)  by  the  sides 
of  the  anirles  and  the  arcs,  the  same  ratio  between  the  sectors 
will  evidently  exist  as  there  is  between  the  arcs.  Wherefore, 
in  equal  circles,  etc. 

Cor.  Hence,  conversely^  arcs  of  the  same  or  equal  circles  will 
have  the  same  ratio  as  the  angles  or  sectors  which  they  measure 
or  subtend — when  greater,  greater ;  less,  less ;  or  equal,  equal. 

Pkop.  XX. — ^Peob. —  Tlie  area  of  a  regular  inscribed  poly ' 
gon^  and  that  of  a  regular  circumscribed  one  of  the  same  nvirv- 
her  of  sides  being  given  y  to  find  the  areas  of  the  regidar  in- 
scribed and  circumscribed  polygons  having  double  the  number 
of  sides. 

].et  A  be  the  center  of  the  circle,  BC  a  side  of  the  inscribed 
polygon,  and  DE  parallel  to  BC,  a  side  of  the  circumscribed 
one.     Draw  the   perpendicular  AFG,  and  the   tangents  BH, 

CK,  and  join  BG;  then  BG  will  be  a  side 
of  the  inscribed  polygon  of  double  the 
number  of  sides;  and  (111.^6,  cor.  l)  IIK  is 
a  side  of  the  similar  cii-cumscribed  one. 
/  \\  //  \  Now,  as  a  like  construction  would  be 
/  x/^  \     Tinule  at  each  of  the  remaining  angles  of 

MAN   ^^Ijp  iwlygon,  it  will   be  sufficient  to  con- 
sider the  i)art  here  represented,  as  the  tri- 
angles connected  with  it  are  evidently  to  each  other  as  the  poly- 
gons of  which  they  are  parts.     For  the  sake  of  brevity,  then. 


n 


G       K 


BOOK  v.]  EUCLID  AND  LEGENDKS.  135 

let  P  denote  the  polygon  whose  side  is  BC,  and  P'  that  whose 
side  is  DE;  and,  in  like  manner,  let  Q  and  Q'  represent  those 
whose  sides  are  BG  and  HK  ;  P  and  P/  therefore,  are  given  ; 
Q  and  Q'  required. 

Now  (V.  ])  the  triangles  ABF,  ABG  are  proportional  to 
their  bases  AF",  AG,  as  they  are  also  to  the  polygons  P,  Q ; 
therefore  AF  :  AG  : :  P  :  Q.  The  triangles  ABG,  ADG  are 
likewise  as  their  bases  AB,  AD,  or  (V.  3)  as  AF,  AG;  and 
they  are  also  as  the  polygons  Q  and  P' ;  therefore  AF  :  AG  : : 
Q  :  P' ;  wherefore  (IV.  7)"  P :  Q  : :  Q  :  P' ;  so  that  Q  is  a  mean 
proportional  between  the  given  polygons  P,  P';  and,  re))re- 
senting  them  by  numbers,  we  have  Q-=PP'',  so  that  the  area 
of  Q  will  be  comjyuted  by  -tnaltlphjing  P  by  P',  and  extracting 
the  square  root  of  the  product. 

Again  :  because  AH  bisects  the  angle  GAD,  and  because 
the  triangles  AHD,  AHG  are  as  their  bases,  Me  have  Gil :  HD 
: :  AG  :  AD,  or  AF  :  AB  : :  AHG  :  AHD.  But  we  have 
already  seen  that  AF  :  AG  ::  P  :  Q;  and  therefore  (IV.  7) 
AHG  :  AHD  : :  P  :  Q.  Hence  (IV.  11)  AUG  :  ADG  : :  P  : 
P-|-Q;  whence,  by  doubling  the  antece- 
dents, 2  AHG  :  ADG  : :  2P  :  P+Q.     But 

D  IT         P         K"         "R 

it  is  evident,  that  whatever  part  the  tri- 


angle ADG  is  of  P^,  the  same  part  of  the 
polyg  )n  Q'  is  the  triangle  AHK,  which  is 
double  of  the  trianojle  AHG,     Hence  the 
last  analogy  becomes  Q'  :  P^ : :  2P  :  P+Q. 
Now  (IV.  2,  cor.  1)  the  product  of  the  ex- 
tremes is  equal  to  the  product  of  the  means;  and  therefore  Q' 
will  be  computed  by  dividing  twice  the  product  o/'P  and  V  by 
P+Q;   and   the   mode   of  finding   Q   has   been   pointed   out 
already. 

Prop.  XXI. — ^Theor. —  Of  regular  polygons  which  have 
eqiial  perimeters.,  that  which  has  the  greater  mimber  of  sides  is 
the  greater. 

Let  AB  be  half  the  sides  of  the  polygon  which  has  the  less 
number  of  sides,  and  BC  a  perpendicular  to  it,  which  will  evi- 
dently pass  through  the  center  of  its  inscribed  or  circumscribed 
circle;  let  C  be  that  center,  and  join  AC.     Then,  ACB  will  be 


136 


THE   ELEMENTS   OF 


[book  V. 


the  angle  at  tlie  center  subtended  by  the  half  side  AB.     ]\I;ike 
BCD  equal  to  the  angle  subtended  at  the  center  of  the  other 
polygon  by  half  its  side,  and  from  C  as  center,  with  CD  as  ra- 
dius, describe  an  arc  cutting  AC  in  E,  and  CB  i)roduced  in  F. 
Then,  it  is  plain,  that  the  angle  ACB  i'*  to  four  right  angles  as 
AB  to  the  common  peiimeter;  and  four  right  angles  are  to 
DCB,  as  the  common  pei-imeter  to  the  half  of 
a  side  of  the  other  polygon,  which,  for  brevi- 
ty, call  S;  then,  ex  mquo^  the  angle  ACB  is 
to  DC]}  as  AB  to  S.     But  (V.  19)  the  angle 
ACB  is  to  DCB  as  the  sector  ECF  to  the 
sector  DCF;  and  consequently  (IV.  1)   the 
sector  ECF  is  to  DCF  as  AB  to  S,  and,  by 
division,  the  sector  ECD  is  to  DCF  as  AB  — 
S  to  S.     Now  the  triangle  ACD  is  greater 
than  the  sector  CED,  and  DCB  is  less  than  DCF.     But  (V.  1) 
these  triangles  are  as  their  bases  AD,  DB  ;  therefore  AD  has  to 
DB  a  greater  i-atio  than  AB  — S  to  S.     Hence  AB,  the  sum  of 
the  first  and  second,  has  to  DB,  the  second,  a  greater  ratio  than 
AB,  the  sum  of  the  third  and  fourth,  has  to  S,  the  fourth  ;  and 
therefore  (IV.  2,  cor.  5)  8  is  greater  than  DB.     Let  then  BG 
be  equal  to  S,  and  draw  GH  parallel  to  DC,  meeting  FC  pro- 
duced in  H.     Then,  since  the  angles  GHB,  DCB  are  equal,  BH 
is  the  perpendicular  drawn  from  the  center  of  the  polygon  hav- 
ing the  greater  number  of  sides  to  one  of  the  sides ;  and  since 
this  is  greater  than   BC,  the  like  perpendicular  in   the  other 
polygon,  while  the  perimeters  are  equal,  it  Ibllows  that  the  area 
of  that  which  has  the  gi-eater  number  of  sides  is  greater  than 
that  of  the  other. 


Prop.  XXIT. — Theor. — If  the  diameter  of  a  circle  1  e  divided 
into  any  two  parts,  AB,  BC,  and  if  semicircles,  ADBjBECy  be 
described  on  opposite  sides  of  these,  the  circle  is  divided  by 
tJieir  arcs  into  ttoo  figures,  GDE,  FED,  the  boundary  of  each  <f 
tchich  is  equal  to  the  circumference  q/'FG  ;  and  which  are  such 
that  AC  :  BC  : :  FG  :  FED,  and  AC  :  AB  ::  ¥G  :  GDE. 

For  (V.  14,  cor.  3)  the  circumferences  of  circles,  and  conse- 
quently the  halves  of  their  circumferences,  are  to  one  another  as 
their  diameters ;  thereiore  AB  is  to  AC  as  the  arc  ADB  to 


BOOK   v.]  EUCLID   AND   LEGENDRE.  137 

AFC,  and  BC  is  to  AC  as  the  arc  BEC  to  AFC.     Hence  (IV. 
1)  AC  is  to  AC  as  the  compound    arc 
ADEC   to    AFC;   therefore  ADEC   is  p 

equal  to  half  the  circumference  ;and  the 
entire  boundaries  of  the  fiirures  GDE, 
FED  are  each  equal  to  the  circumfer- 
ence of  FG. 

Again  (V.  14,  cor,  2):  circles,  and 
consequently  semicircles,  are  to  one  an- 
other as  the  squares  of  their  diameters ; 
therefore  AC'  is  to  AB^  as  the  semicircle 

AFC  to  the  semicircle  ADB.  Hence,  since  (H.  4)  AC'  =  AB» 
-f2AB.BC  +  BC^  we  find  by  conversion  that  AC  is  to  2AB. 
BC  +  BC%  as  the  semicircle  AFC  to  the  remaining  space 
BDAFC;  whence,  by  inversion,  2AB.BC  +  BC'  is  to  AC'  as 
BDAFC  to  the  semicircle  AFC.  But  BC'  is  to  AC=  as  the 
Bemicircle  BEC  to  the  semicircle  AFC;  and  therefore  (IV.  1) 
2AB.BC  + 2BC'  is  to  AC  as  the  compound  figure  FDE  to  the 
semicircle  AFC.  But  (II.  3)  2AB.BC+2BC  =  2AC.BC,  and 
(V.  1)  2AC.BC  :  AC^  :  :  BC  :  ^AC.  Hence  the  preceding 
analogy  becomes  BC  to  ^AC,  as  FED  to  the  semicircle,  or  by 
doubling  the  consequents,  and  by  inversion,  AC  to  BC,  as  FG 
to  FED ;  and  it  would  be  proved,  in  the  same  manner,  that 
AC  :  AB  : :  FG  :  GDE. 

Cor.  Hence  we  can  solve  the  curious  problem,  in  which  it  is 
required  to  divide  a  circle  into  any  proposed  number  of  parts, 
equal  in  area  and  boundary;  as  it  is  only  necessary  to  divide 
the  diameter  into  the  proposed  number  of  equal  parts,  and  to 
desci'ibe  semicircles  on  opposite  sides.  Then,  whatever  part 
AB  is  of  AC,  the  same  part  is  AEG  of  the  circle.  Their  bound- 
aries are  also  equal,  the  boundary  of  each  being  equal  to  the 
circumference  of  the  circle. 

Scho.  Another  solution  would  be  obtained,  if  the  circumfer- 
ence were  divided  into  the  proposed  number  of  equal  parts,  and 
radii  drawn  to  the  points  of  division.  This  division,  hoM'ever, 
can  be  eftected  only  in  some  particular  cases  by  means  of  ele- 
mentary geometry. 

Pkop.  XXIII. — Prob» — To  divide  a  given  circle  ABC  into 


138 


THE   ELEMENTS   OF 


[book  V. 


any  proposed  number  of  equal  parts  by  means  of  concentric 
circles. 

Divide  the  radius  AD  into  the  proposed  number  of  equal 
parts,  suppose  three,  in  the  points  E,  F,  and  through  these 
points  draw  perpendiculars  to  AD,  meeting  a  semicircle  de- 
scribed on  it  as  diameter  in  G,  H; 
from  D  as  center,  at  the  distances  DG, 
DH,  describe  the  circles  GL,  HK ;  their 
circumferences  divide  the  circle  into 
equal  parts. 

Join  All,  DH.  Then  (V.  17,  cor.  2) 
AD,  DH,  DF  being  continual  propor- 
tionals, AD  is  to  DF  as  a  square  de- 
scribed on  AD  is  to  one  described  on 
DH.  But  (V.  14,  cor.  2)  circles  are  proportional  to  the  squares 
of  their  diameters,  and  consequently  to  the  squares  of  their 
radii.  Hence  (IV.  7)  AD  is  to  FD  as  the  circle  ABC  to  the 
circle  HK  ;  and  therefore,  since  FD  is  a  third  of  AD,  HK  is  a 
third  of  ABC.  It  would  be  proved  in  a  similar  manner  that 
AD  is  to  ED  as  ABC  to  GL.  But  ED  is  two  thirds  of  AD, 
and  therefore  GL  is  two  thirds  of  ABC ;  wherefore  the  space 
between  the  circumferences  of  GL,  HK  is  one  third  of  ABC, 
as  is  also  the  remaining  space  between  the  circumferences  of 
ABC  and  GL. 

Cor.  Hence  it  is  plain  that  the  area  of  any  annulus^  or  nng, 
between  the  circumferences  of  two  concentric  circles,  such  as 
that  between  the  circumferences  of  ABC  and  GL  is  to  the  cir- 
cle ABC  as  the  difference  of  the  squares  of  the  radii  DM,  DL 
to  the  square  of  DM ;  or  (II.  5,  cor.  1,  and  HI.  20)  as  the  rect- 
angle AL.LM,  or  the  square  of  the  perpendicular  LB  to  the 
square  of  DM;  and  it  therefore  follows  (V.  14)  that  the  ring  is 
equivalent  to  a  circle  described  with  a  radius  equal  to  LB. 


Prop.  XXIV. — Theor. — If  on  BC  the  hypothenuse  of  a 
ri^/it-angled  triangle  ABC,  a  semicircle,  BAC,  be  described  on 
the  same  side  as  the  triangle,  and  if  semicircles,  ADB,  AEC,  ba 
described  on  the  legs,  falling  without  the  triangle,  the  lunes  or 
crescents  ADB,  AEC,  bounded  by  the  arcs  of  the  semicircles^ 
are  together  equal  to  the  right-angled  triangle  ABC. 


BOOK   v.] 


EUCLID    AND    LEGENDRE. 


139 


For  (V,  14  and  lY.  9)  the  semicircle  ADB  is  to  the  semicircle 
BAG  as  the  square  of  AB  to  the  square  of  BC,  and  AEC  to 
BAG  as  the  square  of  AC  to  the  square  of  BC  ;  whence  (IV.  1) 
the  two  semicircles  ADB,  AEG  taken 
together,  have  to  BAG  the  same  ratio 
as  the  sum  of  the  squares  of  AB,  AG 
to  the  square  of  BC,  that  is,  the  ratio 
of  equality.  From  these  equals  take 
the  segments  AFB,  AGG,  and  there 
remain  the  luncs  DF,  EG  equal  to  the  triangle  ABC. 

Cor.  If  the  legs  AB,  AC  be  equal,  the  arcs  AB'B,  AGG  are 
equal,  and  each  of  them  an  arc  of  a  quadrant ;  also  the  radius 
drawn  from  A  is  perpendicular  to  BC  ;  and  since  the  halves  of 
equals  are  equal,  each  lune  is  equal  to  half  of  the  triangle 
ABC. 

-If,  therefore,  ABC  be  a  qnadi'ant,  and 
on  its  choi'd  a  semicircle  be  described,  the 
lune  comprehended  between  the  circum- 
ferences is  equal  to  the  triangle  ABC ; 
and  since  (II.  13)  a  square  can  be  found 
equal  to  ABC,  we  can  thus  effect  the 
quadrature  of  a  space,  ADC,  bounded 
by  arcs  of  circles. 


Prop.  XXV. — Prob. — To  find  the  area  of  a  circle. 

Sc/io.  1.  The  approximate  area  of  a  circle  can  be  found  by 
means  of  the  twentieth  proposition  of  this  book,  by  what  is 
called  the  method  of  exha'ustions,  giving  an  error  in  excess ; 
viz.,  the  approximate  area  thus  obtained  is  square  of  radius 
multiplied  by  .3.1415926,  etc. 

Geometry  being  an  exact  science,  and  its  conclusions  being 
derived  from  accurate  principles,  the  apj)roximate  area  for  the 
circle  is  not  consistent  with  the  strictness  of  geometrical  rea- 
soning, and  the  area  of  the  circle  must  be  established  exactly 
before  it  can  be  regarded  a  geometrical  truth.  The  reason  why 
the  method  of  e  haustions  gives  the  a]yjyroxim,ate  result,  is  be« 
cause — by  the  twentieth  proposition  of  this  book — the  circum- 
scribed and  inscribed  regular  and  similar  polygons  about  the 
circle  are  supposed^  by  continually  doubling  the  number  of  their 


140  THE   ELEMENTS   OF  [BOOK   V. 

sides,  to  he  made  equivalent  to  the  circle;  but  Carnot,  in  his 
Reflexions  sur  la  Metaphysique  da  Calcul  Infinitesimal^ 
states,  "That  the  ancient  geometers  did 'not  consider  it  con- 
sistent with  the  strictness  of  o-eometrical  reasnTiing  to  re>2:ard 
curve  lines  as  polygons  of  a  great  number  of  sides."  Now, 
the  area  of  any  regular  ])olygon  is  the  rectangle  of  its  a])othem 
and  semi-perimeter;  but  this  area  is  derived  from  the  sixth  cor- 
ollary of  the  twenty-third  proposition  of  the  first  book — since 
it  has  been  shown  in  the  first  corollary  of  the  twentieth  propo- 
sition of  the  same  book,  that  any  rectilineal  figure  can  be 
divided  into  as  many  triangles  as  the  figure  has  sides ;  there- 
fore, in  case  of  a  regular  polygon,  when  triangles  are  formed  ia 
it  by  straight  lines  drawn  from  the  center  to  the  extremities  of 
the  several  sides  of  the  polygon,  the  area  of  the  polygon  be- 
comes by  the  tenth  axiom  of  the  firj^t  book  equivalent  to  the 
sum  of  these  triangles ;  hence  (I.  23,  cor.  6)  each  triangle  is  the 
rectangle  of  the  apothem  of  the  polygon  and  a  semi-side  of  the 
polygon;  therefore  the  area  of  the  polygon  is  (I.  ax.  10)  the 
rectangle  of  its  apothem  and  its  semi-perimeter.  Since  (I.  23, 
C01-.  4)  the  aiea  of  a  triangle  is  derived  from  the  properties  of 
parallel  straight  lives,  and  any  polygon  has  its  sides  straight 
lines  (I.  def  12),  the  pre  perties  o^  parallel  straight  lines  are 
applicable  to  all  polygons;  but  the  circle  being  formed  by  a 
curve  li7ie,  the  properties  of  parallel  straight  lines  are  not  ap- 
plicable to  it ;  hence  the  reason  is  evident  why  the  ancient 
geometers  objected  to  the  curve  line  being  regarded  a  polygon 
of  a  great  number  of  sides.  Euclid,  in  his  ^/e??<e/i^<f,  endeav- 
ored to  sustain  the  proposition,  that  the  circle  is  the  I'cctangle  of 
its  radius  and  semicircumfereiice,  by  what  is  called  the  indirect^ 
apogogic,  or  Meductio  ad  ahsurdwm,  method.  Now,  every 
true  pro])osition  can  be  directly  demonstrated,  and  a  fair  test 
of  the  truth  or  falsity  of  this  proposition  can  be  in  the  success 
or  failure  of  it  being  directly  demonstrated.  I  have  given  the 
d  reel  demonstrations  for  every  other  ])roposition  in  geometry; 
but  I  can  not  do  so  in  this  case — therefore  I  believe  the  proj)0- 
sition  fallacious.  Archimedes  has  shown  that  the  relation  of 
diameter  to  the  circumference  of  a  circle  expressed  in  numbers, 
to  be  as  7  to  22 — which  is  practically  correct.  Among  isoperi- 
metrical  figures,  the  circle  contains  the  greatest  area ;  there- 


BOOK  v.]  EUCLID  AND  LEGENDRE.  141 

fore  when  22  expresses  the  circumference  of  a  circle,  the  perim- 
eter of  its  equivalent  square  must  he  greater  than  22;  and  if  a 
cube  be  inechauically  constructed  upon  a  base  whose  perimeter 
is  24.2487  +  ,  it  will  be  equivalent  to  a  cylinder  of  same  height, 
the  diameter  of  whose  base  is  7. 

Now,  when  24.2487+  expresses  the  perimeter  of  a  square, 
each  of  its  sides  (I.  23,  cor.  1)  will  be  6.0621  +  ;  and  its  area 
■will  be  36.75,  or  three  times  square  of  the  radius  of  the  circle. 
Hence  we  get  by  mechanical  construction  less  than  what  is  ob- 
tained by  the  method  of  exhaustions.  The  geometrical  con- 
firmation of  the  mechanical  construction  is  given  in  the  second 
corollary  to  the  seventeenth  proposition  of  the  sixth  book. 

Scho.  2.  Euclid  has  endeavored  to  demonstrate  that  the  cir- 
cle is  the  rectangle  of  circumference  ajid  semi-i'adius.  Now, 
the  square  equal  to  circle  is  somewhere  between  the  inscribed 
and  circumscribed  squares,  and  its  area  is  equal  to  its  perime- 
ter multiplied  by  less  than  semi-radius;  consequently  the  rect- 
angle of  circumference  and  semi-radius  will  produce  more  than 
area  of  circle. 

{^Or  Thomson^ s  Eficlid,  Appendix,  DooJc  I.,  Prop.  JTJTXZX) 

"  The  area  of  a  circle  is  equal  to  the  rectangle  under  its  ra- 
dius, and  a  straight  line  equal  to  half  its  circumference.  Let 
AB  be  the  radius  of  the  circle  BC  ;  the  area  of  BC  is  equal  to 
the  rectangle  under  AB  and  a  straight  line  D  equal  to  half  the 
circumference. 

"  For  if  the  rectangle  AB.  D  be  not  equal  to  the  circle  BC,  it 
is  equal  to  a  circle  either  greater  or  less  than  BC.  First,  sup- 
pose, if  possible,  the  rectangle  AB.  D  to  be  the  area  of  a  circle 
EF,  of  which  the  radius  AE  is  greater  than  AB."  Here  Euclid 
is  inconsistent  with  his  own  proposition  :  at  the  very  stait  he 
bases  his  argument  upon  a  contradiction.  He  premises  that 
the  area  of  a  circle  is  equal  to  the  rectangle  under  its  radius, 
and  a  straight  line  equal  to  half  its  circumference  ;  then  sujy- 
pose,  i.  6.,  asks  to  be  granted  for  the  sake  of  argument,  that 
that  same  rectangle  is  equal  to  a  larger  circle.  Why  does  he 
resort  to  this  subterfuge?  It  will  be  said  to  show  the  Heductio 
ad  absurduni  ;  very  well,  let  us  follow  his  argument :  "and  let 
GHK  be  a  regular  polygon  described  about  the  circle  BC,  such 


142  THE   ELKMENT8   OF  [bOOK  V. 

that  its  sides  flo  not  meet  the  circumference  of  EF.  Then,  by 
dividing  this  polygon  into  triangles  by  radii  drawn  to  G,  H, 
K,  etc.,  it  would  be  seen  that  its  area  is  equal  to  the  rectangle 
under  AB  and  half  its  perimeter.  But  the  perimeter  of  the 
polygon  is  greater  than  the  ciixiumference  of  BC,  and  therefore 
the  area  of  the  polygon  is  greater  than  the  rectangle  AB.  D  ; 
that  is,  by  hypothesis,  than  the  area  of  the  circle  EF,  which  is 
absurd."  What  is  absurd  ?  That  the  circle  EF  is  greater  than 
the  polygon  GHK,  etc.,  or  Euclid's  argument?  The  absurdity 
is  in  considering  the  area  of  a  circle  equal  to  a  larger  circle. 
An  argument  based  upon  absurdity  must  necessarily  lead  to 
absurdity,  which  in  fact  has  been  the  case.  When  Euclid  sup- 
posed, i.  e.,  asked  to  be  granted  for  the  cake  of  argument,  AB. 
D= circle  EF,  it  does  not  prove  the  area  of  polygon  greater 
than  the  area  of  circle  EF,  because  he  at  the  start  supposed 
AB.  D  =  circle  EF,  and  consistently  with  his  hypothesis  and 
his  argument,  it  must  be  so  to  the  end  ;  therefore,  consistently 
with  his  argument  and  his  hypothesis,  AB.  D  is  greater  than 
the  area  of  polygon  GHK,  etc.  The  first  part  of  Euclid's  prop- 
osition is  nothing  more  than  a  demonstration  to  prove  the  area 
of  a  circle  is  greater  than  the  area  of  a  polygon  drawn  within 
the  circle.  And  the  second  part  of  Euclid's  proposition  is 
nothing  more  than  a  demonstration  to  prove  a  circle  less  than 
the  circumscribing  polygon.  This  proposition  of  Euclid  is 
very  sophistical,  and  consequently  its  fallacy  has  been  imde- 
tected,  owing  no  doubt  to  the  repute  of  Euclid,  and  to  the  sup- 
position that  Euclid  argued  from  axioms,  and  consistently  with 
the  principles  of  geometry,  which  he  did  ;  but  in  this  instance 
he  deceived  himself,  and  consequently  all  those  who  believe 
him  the  oracle  of  geometry.  When  he  attempted  to  prove 
AB.  D=area  of  circle  BC,  it  was  contradictory  to  his  argument 
to  suppose  AB.'D  =  area  of  circle  EI*';  because  when  he  based 
his  argument  upon  the  premies  that  AB.  D=:area  of  circle  EF, 
consistency  demanded  that  he  should  stand  by  his  premise,  and 
not  forsake  it  as  soon  as  it  led  to  an  absurdity,  and  judge  a  cir- 
cle less  than  a  polygon  within  a  cii'cle.  Tlie  absurdity  is  in 
his  own  argument,  to  base  it  upon  a  supposition  which  he  knew 
was  inconsistent  with  his  proposition,  and  the  inconsistence  to 
drop  his  premiss  when  he  perceived  it  led  to  an  absurdity ;  as 


BOOK  v.]  EUCLID  AND  LEGENDRE.  143 

AB.  D  is  less  than  the  circle  EF,  it  is  a  very  fallacious  argu- 
ment, when  based  on  the  supposition  that  they  are  equal,  and 
it  leads  to  an  absurdity ;  and  very  inconsistent  with  geometri- 
cal reasoning  for  Euclid  to  drop  at  the  conclusion  of  his  argu- 
ment the  very  premiss  upon  which  he  based  his  argument. 
Every  method  of  demonstration,  as  well  as  that  method  termed 
Heductio  adalsurdttm,  require  that  the  premiss  which  is  adopt- 
ed at  the  start  be  retained  to  the  conclusion.  And  when  Euclid 
adopted  AB.  D=circle  EF  at  the  commencement  of  his  demon- 
stration, consistence  of  reason  and  science  demanded  that  he 
fihould  have  kept  it  to  the  conclusion,  and  then  there  would 
have  been  no  absurdity,  but  a  demonstration  to  prove  that  the 
polygon  GHK,  etc.,  is  less  than  circle  EF.  But  Euclid  had 
in  his  mind  AB.  D  =  circle  BC ;  forgetting  that  he  had  adopt- 
ed AB.  D= circle  EF,  and  had  stiU  to  prove  AB.  D= circle 
BC. 


END   OF  BOOK   FIFTH. 


BOOK    SIXTH. 
ON  THE  PLANE  AND  S0LID3. 

DEFINITIONS. 

1.  A  STRAIGHT  line  is  said  to  be  perpendicular  to  a  plane 
when  it  makes  ri'jrht  ansjles  with  all  straiiiht  lines  meeting'  it  iu 
that  plane. 

2.  The  inclination  of  two  planes  which  meet  one  another  is 
the  angle  contained  by  two  straight  lines  drawn  from  any  point 
of  their  common  section  at  right  angles  to  it,  one  upon  each 
plane.  The  angle  which  one  plane  makes  >.ith  another  is 
Boraetimes  called  a  dihedral  angle. 

3.  If  that  angle  be  a  right  angle,  the  planes  arc  perpendicu- 
lar to  one  another. 

4.  Parallel  planes  are  such  as  do  not  meet  one  another, 
though  produced  ever  so  far  in  every  direction. 

5.  A  solid  angle  is  that  which  is  made  by  more  than  two 
plane  angles  meeting  in  one  point,  and  not  lying  in  the  same 
plane. 

If  the  number  of  plane  angles  be  three,  the  solid  angle  is  tri- 
hedral;  if  four,  tetrahedral ;  if  more  than  io\x\\  polyhedral. 

6.  Kpolyfiedron  is  a  solid  figure  contained  by  plane  figures. 

If  it  be  contained  by  four  plane  figures,  it  is  called  a  tetrahe- 
dron ;  if  by  six,  a  hexahedron  ;  if  by  eight,  an  octahedron  ;  if 
by  twelve,  a  dodecahedron  ;  if  by  twenty,  an  icosahedron^  etc. 

1.  A  regular  body^  or  regular  polyhedron^  is  a  solid  con- 
tained by  plane  figures,  which  are  all  equal  and  similar. 

8.  Of  solid  figures  contained  by  planes^  those  are  similar 
which  have  all  their  solid  angles  equal,  each  to  each,  and  which 
are  contained  by  the  same  number  of  similar  plane  figures,  simi- 
larly situated. 

9.  A  pyra^nid  is  a  solid  figure  contained  by  one  plane  figure 
called  its  base,  and  by  three  or  more  triangles  meeting  in  a 
point  without  the  plane,  called  the  vertex  of  the  pyramid. 


BOOK   VI.]  EUCLID   AND   LEGENDEE.  145 

10.  A ]jrism  is  a  solid  fijjiire,  the  ends  or  hases  of  \vliich  are 
parallel,  and  are  equal  and  similar  plane  figures,  and  its  otlier 
boundaries  are  parallelograms.  One  of  tliese  parallelograms 
also  is  sometimes  regarded  as  the  base  of  the  j)rism. 

11.  Pyramids  and  prisms  are  said  to  be  triangular  wlien 
their  bases  are  triangles ;  quadrangular,  when  their  bases  are 
quadrilaterals ;  pentagonal,  when  ihey  are  pentagons,  etc. 

12.  The  altitude  of  a  pyramid  is  the  perpendicular  drawn 
from  its  vertex  to  its  base;  and  the  altitude  of  a  prism  is  either 
a  perpendicular  drawn  from  any  point  in  one  of  its  ends  or 
bases,  to  the  other;  or  a  perpendicular  to  one  of  its  bounding 
parallelograms  from  a  point  in  the  line  opposite.  The  first  of 
these  altitudes  is  sometimes  called  the  length  of  the  prism. 

13.  A  prism,  of  which  the  ends  or  bases  are  perpendicular  to 
the  other  sides,  is  called  a  right  pris7n  y  any  other  is  an  ohluiue 
pristn. 

14.  A  parallelopiped  is  a  prism  of  which  the  bases  are  par- 
allelograms. 

15.  A  parallelopiped  of  which  the  bases  and  the  other  sides 
are  rectangles,  is  said  to  be  rectangular. 

16.  A  cube  is  a  rectangular  parallelopiped,  which  has  all  its 
six  sides  squares. 

17.  A  sphere  is  a  solid  figure  described  by  the  revolution  of 
a  semicircle  about  its  diameter,  which  remains  unmoved. 

18.  The  axis  of  a  sphere  is  the  fixed  straight  line  about 
which  the  semicircle  revolves. 

19.  The  center  of  a  sphere  is  the  same  as  that  of  the  generat- 
ing semicircle. 

20.  A  diameter  of  a  sphere  is  any  straight  line  which  passes 
through  the  center,  and  is  terminated  both  ways  by  its  surface. 

21.  A  cone  is  a  solid  figure  described  by  the  revolution  of  a 
riirht-aniiled  triangle  about  one  of  the  legs,  which  remains  fixed. 

If  the  fixed  leg  be  equal  to  the  other  leg,  the  cone  is  called 
a  right-angled  cone ;  if  it  be  less  than  the  other  leg,  an  obtuse- 
angled,  and  if  greater,  an  acute-angled  cone. 

22.  The  axis  of  a  cone  is  the  fixed  straight  line  about  which 
the  triangle  revolves. 

23.  The  base  of  a  cone  is  the  circle  described  by  the  leg 
■which  revolves. 

10 


146 


THE   ELEMENTS   OF 


[book  VI. 


24.  A  cylinder  is  a  solid  figure  described  by  the  revolution 
of  a  rectangle  about  one  of  its  sides,  which  remains  fixed. 

25.  The  axis  of  a  cylinder  is  the  fixed  straight  line  about 
which  tlie  rectano-le  revolves. 

26.  The  bases  or  ends  of  a  cylinder  are  the  circles  described 
by  the  two  revolving  opposite  sides  of  the  rectangle. 

27.  Similar  cones  and  cylinders  are  those  which  have  their 
axes  and  the  diameters  of  their  bases  proportionals. 

PROPOSITIONS. 

Prop.  I. — Theor. —  One  part  of  a  straight  line  can  not  be  in 
a  plane  and  another  part  above  it. 

J^et  EFGH  be  a  plane,  then  the  straight  line  AB  will  be 

wholly   in  the  plane.     By  def  1, 

Book  VI.,  and  def  7,  Book  L,  AB, 

being  a  straight  line  in  the  plane 

EFGH,  is   wholly  in   that  plane, 

and  can  not  have  one  part  in  the 

plane  and  another  part  above  it. 

Cor.  1.  Hence  two  straight  lines 

which  cut   one  another  are  in   the   same  plane ;  so  also  are 

three  straight  lines  which  meet  one  another,  not  in  the  same 

point. 

Cor.  2.  Hence,  if  two  planes  cut  one  another,  their  common 
section  is  a  straight  line. 


Prop.  H. — Theor, — If  a  straight  li?7£  be  perpendicular  to 
each  of  two  straight  lines  at  their  point  of  intersection^  it  is 
also  perpendicvlar  to  the  plane  in  which  they  are. 

Let  the  straight  line  EF  be  perpendicular  to  each  of  the 
straight  lines  AB,  CD  at  their  intersection  E ;  EF  is  also  per- 
pendicular to  the  plane  passing  through  AB,  CD. 

T.ake  the  straight  lines  EB,  EC  equal  to  one  another,  and 
join  BC ;  in  BC  and  EP"  take  any  points  G  and  F,  and  join 
EG,  FB,  FG,  FC.  Then,  in  the  triangles  BEF,  CEF,  BE  is 
equal  to  CE;  EF  common;  and  the  angles  BEF,  CEF  are 
equal,  being  (hyp.)  right  angles;  therefore  (I.  3)  BF  is  equal 
to  CF.     The  triangle  BFC  is  therefore   isosceles ;  and  (IL  5, 


BOOK    VI.] 


EUCLID    AND   LKGENDRE. 


IIT 


cor,  5)  the  square  of  BF  is  equivalent  to  the  square  of  FG  and  the 
rectangle  BG.GC.  Foi"  the  same  reason,  because  (const.)  the 
triangle  BEG  is  isosceles,  the  square  of  BE 
is  equivalent  to  the  square  GE  and  the  rect- 
angle BG.GC.  To  each  of  these  add  the 
square  of  EF ;  then  tlie  squaies  of  BE,  EF 
are  equivalent  to  the  squares  of  GE,  EF, 
and  the  rectangle  BG.GC,  But  (T.  24,  cor, 
1)  the  squares  of  BE,  EFare  equivalent  to 
the  square  of  V>F,  because  BEF  is  a  right 
angle;  and  it  has  been  shown  that  the  square  of  BF  is  equiva- 
lent to  the  .square  of  FG  and  the  rectangle  BG.GC;  therefore 
the  square  of  FG  and  the  rectangle  BG.GC  are  equivalent  to 
the  squares  of  GE,  EF,  and  the  rectangle  BG.GC.  Take  the 
rectangle  BG.GC  from  each,  and  there  remains  the  square  of 
FG,  equivalent  to  the  squares  of  GE,  EF  ;  wherefore  (I.  24,  cor.) 
FEG  is  a  right  angle.  In  the  same  manner  it  would  be  proved 
that  EF  is  perpendicular  to  any  other  straight  line  drawn 
through  E  in  ihe  plane  passing  through  AB,  CD.  But  (VI. 
def  1)  a  straight  line  is  perpendicular  to  a  plane  when  it  makes 
right  angles  with  all  straight  lines  meeting  it  in  that  plane; 
therefore  EF  is  perpendicular  to  the  plane  of  AB,  CD.  Where- 
fore, if  a  straight  line,  etc. 

Cor.  Hence  (VI.  def.  l)  if  three  straight  lines  meet  all  in 
one  point,  and  a  straight  line  be  perpendicular  to  each  of  them 
at  that  point,  the  three  straight  lines  are  in  the  same  plane, 

Pkop,  III. — Theor,  — Tf  tioo  straight  lines  he  perpendicular 
to  the  same  plane,  they  are  parallel  to  one  another. 

Let  the  straight  lines  AB,  CD  be  at  right  angles  to  the  same 
plane  BDE;  AB  is  parallel  to  CD. 

Let  them  meet  the  plane  in  the  points  B, 
D  ;  join  BD,  and  draw  DE  perpendicular  to 
BD  in  the  plane  BDE  ;  make  DE  equal  to 
AB,  and  join  BE,  AE,  AD.  Then,  because 
AB  is  perpendicular  to  the  plane,  each  of  the 
angles  ABD,  ABE  is  (VL  def  1)  a  right  an- 
gle. For  the  same  reason,  CDB,  CDE  are 
right  angles.     And  because  AB  is  equal  to 


148 


THE   ELKMENTS    OF 


[lIOOK   VI. 


DE,  BD  common,  and  the  angle  ABD  equal  to  BDE,  AD  is 
equal  (I.  3)  to  DE. 

A-^ain  :  in  the  triangles  ABE,  ADE,  AB  is  equal  to  DE,  BE 
to  AD,  and  AE  common ;  therefore  (I.  4)  the  angle  ABE  is 
equal  to  EDA ;  but  ABE  is  a  right  angle ;  therefore  EDA  is 
also  a  right  angle,  and  ED  perpendicular  to  DA ;  it  is  also 
perpendicular  to  each  of  the  two  BD,  DC;  therefore  (VI.  2, 
cor.)  these  three  straight  lines  DA,  DB,  DC  are  all  in  the  same 
plane.  But  (VI.  1,  cor.  1)  AB  is  in  the  plane  in  which  are  BD, 
DA  ;  therefore  AB,  BD,  DC  are  in  one  plane.  Now  (hyp.) 
each  of  the  angles  ABD,  BDC  is  a  right  angle ;  theiefore  (L 
16,  cor.  1)  AB  is  parallel  to  CD.     Wherefore,  etc. 

Cor.  1.  Hence  (I.  def.  11)  if  two  straight  lin^s  be  parallel, 
the  straight  line  drawn  from  any  point  in  the  one  to  any  point 
in  the  other  is  in  the  same  plane  with  the  parallels. 

Cor.  2.  Hence,  also,  if  one  of  two  parallel  straight  lines  be 
perpendicular  to  a  plane,  the  other  is  also  perpendicular  to  it. 

Also,  two  straight  lines  which  are  each  of  them  T)arall(l  to 
the  same  straight  line,  and  are  not  both  in  the  same  plane  with 
it,  are  parallel  to  one  another. 

Scho.  The  same  has  been  proved  (I.  1 7)  respecting  straight 
lines  in  the  same  plane;  therefore,  universally,  straight  hues 
•which  are  parallel  to  the  same  straight  line,  are  parallel  to  one 
another. 


Prop.  TV. — Tfieor. — If  two^  straight  lines  meeting  one  an- 
other be  parallel  to  two  others  that  meet  one  another^  and  are 
not  in  the  same  plane  with  the  first  two  ;  tlie  first  two  and  the 
other  two  contain  equal  angles. 

Let  the  straight  lines  AB,  BC,  which  meet  one  another,  be 
parallel  to  DE,  EF,  which  also  meet  one 
another,  but  are  not  in  tlie  same  ])lane 
with  AB,  BC  ;  the  angle  ABC  is  equal  to 
DEF. 

Take  BA,  BC,  ED,  EF,  all  equal  to  one 
another,  and  join  AD,  CF,  BE,  AC,  DF. 
Because  BA  is  equal  and  parallel  to  ED, 
thereft)re  AD  is  (I.  15,  cor.  l)  both  equal 
and  parallel  to  BE.     For  the   same  reason,  CF  is  equal  and 


^ 

N 

E 

\ 

D 


f 


BOOK  VI.] 


EUCLID  AND  LEGKNDKK. 


149 


parallel  to  BE.  Therefore  AD  and  CF  Ix'ing  each  of  them 
parallel  to  BE,  are  (VI.  3,  cor,  2)  parallel  to  one  another. 
They  are  also  (I.  ax,  1)  equal;  and  AC,  DF  join  them  toward 
the  same  parts;  and  therefore  (I.  15,  cor.  1)  AC  is  equal  and 
parallel  to  DF.  And  because  AB,  BC  are  equal  to  DE,  EF, 
and  AC  to  DF,  the  angle  ABC  is  equal  (I.  4)  to  DEF.  There- 
fore, if  two  stiMight  line?,  etc. 

Schu.  Or  supplemental  ones,  as  will  be  plain  after  the  de- 
monstration here  given,  if  AB  be  produced  through  15.  This 
generalizes  the  third  corollary  to  the  sixteenth  proposition  of 
the  first  book. 


Prop.  V. — Prob. —  To  clrmo  a  straight  line  perpendicular  to 
a  plane,  from  a  given  point  above  it. 

Let  A  be  the  given  point  above  the  plane  BII ;  it  is  required 
to  draw  from  A  a  perpendicular  to  BH. 

In  the  plane  draw  any  straight  line  BC,  and  (I.  8)  from  A 
draw  AD  perpendicular  to  BC.  Then,  if  AD  be  also  perpen- 
dicular to  the  plane  BH,  the  thing  required  is  done.  But  if  it 
be  not,  from  D  (I.  7)  draw  DE, 
in  the  i)lane  BII,  at  right  angles 
to  BC  ;  from  A  draw  AF  perpen- 
dicular to  DE;  and  through  F 
draw  (I.  18)  Gil  parallel  to^  BC. 
Then,  because  BC  is  at  right  an- 
gles to  ED  and  DA,  BC  is  at 
right  angles  (VI.  2)  to  the  plane 
passing   through    ED,  DA ;  and 

GH  being  parallel  to  BC,  is  also  (VT.  3,  cor.  2)  at  right  angles 
to  the  plane  through  ED,  DA  ;  and  it  is  therefore  perjjendicular 
(VI.  def  1)  to  every  straight  line  meeting  it  in  that  ])lane;  GM 
is  consequently  perpendicular  to  AF.  Therefore  AF  is  per- 
pendicular to  each  of  the  straight  lines  Gil,  DE ;  and  conse- 
quently (VI.  2)  to  the  plane  BHj  wherefore  AF  is  the  perpen- 
dicular required. 


Prop.  VI. — ^Prob.  —  To  draw  a  straight  line  perpendicular  to 
a  given  plane  from  a  point  given  in  the  plane. 


150 


THE   ELEMENTS   OF 


[book  VI. 


D 


B 


Let  A  be  the  point  given  in  the  plane  ;  it  is  required  to  draw 
a  perpendicular  from  A  to  the  plane. 

From  any  point  B,  above  the  plane,  draw  (VI.  5)  BC  per- 
pendicular to  it ;  if  this  pass  through  A,  it  is 
the   perpendicular  required.     If  not,   from   A 
draw  (I.  18)  AD  parallel   to  BC.     Then,  be- 
cause AD,  CB  are  parallel,  and  one  of  them, 
BC,  is  at  right  angles  to  tlie  given  plane,  the 
other,  AD,  is  also  (VI.  3,  cor.  2)  at  right  angles 
to  it. 
Scho.  From  the  same  point  in  a  given  plane  there  can  not 
be  two  straight  lines  drawn  perpendicular  to  the  plane  upon 
the  same  side  of  it ;  and  there  can  be  but  one  perpendicular  to 
a  plane  from  a  point  above  it. 

Cor.  Hence  planes  to  which  the  same  straight  line  is  perpen- 
dicular, are  i^arallel  to  one  another. 


E 


Prop.  VII. — Theor. —  Two  planes  are  parallel^  if  tico 
straiglit  lines  which  meet  one  another  on  one  of  them  be  parallel 
to  two  which  meet  on  the  other. 

Let  the  straight  lines  AB,  BC  meet  on  the  plane  AC,  and 

DE,  EF  on  the  plane  DF ;  if  AB,  BC  be  parallel  to  DE,  EF, 
the  plane  AC  is  parallel  to  DF. 

From  B  draw  (VI.  3,  cor.  2)  BG  perpendicular  to  the  plane 

DF,  and  let  it  meet  that  plane  in  G ;  and  through  G  draw  (L 

18)  Gil  parallel  to  ED,  and  GK  to 
EF.  Then,  because  BG  is  perpendicu- 
lar to  the  plane  DF,  each  of  the  angles 
BGII,  BGK  is  (VI.  def  1)  a  right  an- 
gle ;  and  because  (VI.  3,  cor.  2)  BA  is 
parallel  to  Gil,  each  of  thom  being  par- 
allel to  DE,  the  angles  GBA,  BGII  are 
together  equal  (1. 1 6,  cor.  1 )  to  two  right 

angles.  But  BGII  is  a  risjht  aniile ; 
therefore,  also,  GBA  is  a  right  angle,  and  GB  perpendicular  to 
BA.  For  the  same  reason,  (iB  is  pei-pendicular  to  BC.  Since, 
therefore,  GB  is  perpendicular  to  BA,  BC,  it  is  perpendicular 
(VL  2)  to  the  plane  AC;  and  (const.)  it  is  perpendicular  to  the 
plane  DF.     But  (VI.  6,  cor.)  planes  to  which  the  same  straight 


BOOK   VI.] 


EUCLID    AND   LEGENDRE. 


151 


line  is  perpendicular  are  parallel  to  one  another;  therefore  the 
planes  AC,  DF  are  parallel.     Wherefore,  two  planes,  etc. 

Cor.  1.  Hence,  if  two  parallel  planes  be  cut  by  another  plane, 
their  common  sections  with  it  are  parallels. 

Cor.  2.  If  a  straight  line  be  perpendicular  to  a  plane,  every 
plane  which  passes  through  it  is  perpendicular  to  that  plane. 

C-yr.  3.  Hence,  if  two  planes  cutting  one  another  be  each 
perpendicular  to  a  third  plane,  their  common  sectiiBU  is  perpen- 
dicular to  the  same  plane. 


Paop.  VIH. — Theor, — If  two  straight  lines  be  cut  by  parallel 
planes,  they  are  cut  in  the  same  ratio. 

Let  the  straight  lines  AB,  CD  be  cut  by  the  parallel  planes 
GH,  KL,  MN,  in  the  points  A,  E,  B  ;  C,  F,  D;  as  AE  :  EB  :  : 
CF  :  FD. 

Join  AC,  BD,  AD,  and  let  AD  meet  KL 
in  X ;  join  also  EX,  XF.  Because  the 
two  parallel  planes  KL,  MN  are  cut  by 
the  plane  EBDX,  the  common  sections 
EX,  BD  are  (VL  7,  cor.)  parallel.  For 
the  same  reason,  because  GH,  KL  are  cut 
by  the  plane  AXFC,  the  common  sections 
AC,  XF  are  parallel.  Then  (V.  2)  because 
EX  is  parallel  to  BD,  a  side  of  the  triangle 
ABD,  AE  :  EB  : :  AX  :  XD ;  and  be-  jf 
cause  XF  is  parallel  to  AC,  a  side  of  the  triangle  ADC,  AX  : 
XD  :  :  CF  :  FD ;  and  it  was  proved  that  AX  :  XD  :  :  AE  : 
EB  ;  therefore  (IV.  7)  AE  :  EB  :  :  CF  :  FD.  Wherefore,  if 
two  straight  lines,  etc. 


Prop.  IX — Theor. — If  a  solid  angle  be  contained  by  three 
plane  angles,  any  two  of  them,  are  greater  than  the  tliird. 

Let  the  solid  angle  at  A  be  contained  by  the  three  plane 
angles  BAC,  CAD,  DAB ;  any  two  of  these  are  greater  than 
the  third. 

If  the  angles  be  all  equal,  it  is  evident  that  any  two  of  them 
are  greater  than  the  third.  But  if  they  be  not,  let  BAC  be 
that  angle  which  is  not  less  than  either  of  the  other  two,  and 


153 


THE    ELEMENTS    OF 


[book   VI. 


is  rrreator  than  one  of  thorn,  DAB;  aiul  make  in  tlie  plane  of 
BA,  AC  the  angle  BAE  equal  (I.  13)  to  DAB;  make  AE 
equal  to  AD;  through  E  draw  BEC  cutting  AB,  AC  in  the 
points  B,  C,  and  join  DB,  DC.  Then,  in  tlie  triangles  BAD, 
BAE,  because  DA  is  equal  to  AE,  AB  common,  and  the  angle 
DAB  is  equal  to  EAB,  DB  is  equal  (I.  3)  to  BE.  Again  : 
because  (I.  2],  cor.)  BD,  I  C  are  greater  than  CB,  and  one  of 
them,  BD,  has  been  pioved  equal  to  BE,  a  part  of  CB,  there- 
fore the  other,  DC,  is  greater  (I.  ax.  5)  than  the  remaining  part, 
EC.  Then,  because  DA  is  equal  (const.)  to 
AE,  and  AC  common,  but  the  base  DC 
greater  than  the  base  EC,  therefore  (T.  21) 
the  angle  DAC  is  greater  than  EAC,  and 
(const.)  the  angles  DAB,  BAE  are  equal ; 
Avherefore  (I.  ax.  4)  the  angles  DAB,  DAC 
are  together  greater  than  BAE,  EAC,  that  is, 
than  BAC.  But  BAC  is  not  less  than  either 
of  the  angles  DAB,  DAC  ;  therefore  BAC, 
with  either  of  them,  is  greater  than  the 
other.     Wherefore,  if  a  solid  angle,  etc. 

Cor.  1.  If  every  two  of  three  plane  angles  be  greater  than 
the  third,  and  if  the  straight  lines  which  contain  them  be  all 
equal,  a  ti-iangle  may  be  made,  having  its  sides  equal,  each  to 
each,  to  the  straight  lines  that  join  the  extremities  of  those 
equal  straight  lines. 

Cor.  2.  If  two  solid  angles  be  each  contained  by  three  plane 
angles,  equal  to  one  another,  each  to  each  ;  the  planes  in  which 
the  equal  angles  are,  have  the  same  inclination. 

Cor.  3.  Two  solid  angles,  contained  each  by  three  plane  an- 
gles which  are  equal  to  one  another,  each  to  each,  and  alike 
situated,  are  equal  to  one  another. 

Cor.  4.  Solid  figures  contained  by  the  same  number  of  equal 
and  similar  planes,  alike  situated,  and  having  none  of  their 
solid  angles  contained  by  more  than  three  plane  angles,  are 
equal  and  similar  to  one  another. 

Cor.  5.  If  a  solid  be  contained  by  six  planes,  two  and  two 
of  which  are  parallel,  the  opposite  planes  are  similar  and  equal 
parallelograms. 


BOOK  VI.]         EUCLID  AND  LEGENDKE.  153 

Prop.  X. — Theor. — Ecery  solid  angle  is  contained  hy  plane 
angles^  which  are  together  less  than  four  right  angles. 

Let  the  solid  angle  <it  A  be  contained  hy  any  nnmher  of 
plane  angles,  BAC,^  CAD,  DAE,  EAF,  FAB ;  these  together 
are  less  than  four  right  angles. 

Let  the  planes  in  which  the  angles  are  be  cut  by  a  plane, 
and  let  the  connnon  sections  of  it  with  those  planes  be  BC,  CD, 
DE,  EF,  FB.  Then,  because  the  solid  angle  at  W  is  contained 
by  three  plane  angles,  CBA,  ABF,  FBC,  of  which  (VL  9)  any 
two  are  greater  than  the  third,  CBA,  ABF  are  greater  than 
FBC.  For  the  same  reason,  the  two  plane  angles  at  each  of 
the  ])oints  C,  D,  E,  F,  viz.,  the  angles  which  are  at  the  bases 
of  the  triangles  having  the  common  vertex  A,  are  greater  than 
the  third  angle  at  the  same  point,  which  is 
one  of  the  ansfles  of  the  figure  BCDEF. 
Therefore  all  the  angles  at  the  bases  of  the 
triangles  are  together  greater  than  all  the 
angles  of  that  figure;  and  because  (L  20) 
all  the  anoles  of  the  triangles  are  together 
equal  to  twice  as  many  right  angles  as  there 
are  triangles — that  is,  as  there  are  sides  in 
the  figure  BCDEF;  and  that  (L  20,  cor.  1)  all  the  angles  of  the 
figure,  together  with  four  right  angles,  are  likewise  equal  to 
twice  as  many  riuht  angles  as  there  are  sides  in  thefi<»-ure; 
therefore  all  the  angles  of  the  triangle  are  equal  to  all  the 
angles  of  the  figure,  together  with  four  right  angles.  But  all 
the  angles  at  the  bases  of  the  triangles  are  greater  than  all  the 
angles  of  the  figure,  as  has  been  proved  ;  wherefore  the  re- 
maining angles  of  the  triangles,  viz.,  those  at  the  vertex,  wiiich 
contain  the  solid  angle  at  A,  are  less  than  four  ri^-ht  angles. 
Therefore,  every  solid  angle,  etc. 

Scho.  This  proposition  does  not  necessarily  hold,  if  any  of 
the  angles  of  the  rectilineal  figure  BCDEF  be  re-entrant;  or, 
■which  is  the  same,  if  any  of  the  planes  foi-ming  the  solid  angle 
at  A,  being  produced,  pass  through  that  angle. 

Prop.  XI. — Prob. —  To  make  a  solid  angle  having  the  angles 
containing  it  equal  to  three  given  jL>la?ie  angles^  any  two  of 


154 


THE   ELEMENTS   OF 


[book  VI. 


which  are  greater  than  the  third^  and  all  three  together  less  than 
four  right  angles. 

Let  B,  E,  H  be  given  plane  angles,  any  two  of  which  are 
greater  than  the  third,  and  all  of  them  together  less  than  four 
right  angles;  it  is  required  to  make  a  solid  angle  contained  by 
plane  angles  equal  to  B,  E,  H,  each  to  each. 

Fjom  the  lines  containing  the  angles,  cut  oif  BA,  BC,  ED, 
EF,  IIG,  HK,  all  equal  to  one  another,  and  join  AC,  DF,  GK ; 
then  (VI.  9,  cor.  1)  a  triangle  may  be  made  of  three  straight 
lines  equal  to  AC,  DF,  GK.  Let  this  (L  12)  be  the  triangle 
LMN,  AC  being  equal  to  LM,  DF  to  MN",  and  GK  to  LN. 

About  LiMN  describe  (IIL  25,  cor.  2)  a  circle,  and  draw  the 
radii,  LO,  JMO,  NO ;  draw  also  OP  (VI.  6,  cor.)  perpendicu- 
lar to  the  plane  LMN.  Then,  any  of  the  radii  LO,  MO,  NO  is 
less  than  AB.  Find  (I.  24,  cor.  3)  the  side  of  a  square  equiva- 
lent to  the  difference  of  the  squares  of  AB  and  LO  ;  make  OP 
equal  to  that  side,  and  join  PL,  PM,  PN  ;  the  plane  angles 
LPM,  MPN,  and  NPL  form  the  solid  angle  required. 

For,  since  OP  is  (const.)  perpendicndar  to  the  ])lane  LIVIN, 
the  angles  LOP,  MOP,  NOP  are  (VI.  def  1)  right  angles; 
and  therefore,  since  in  the  triangles  OLP,  OMP,  ONP  the 
sides  OL,  OM,  ON  are  equal,  OP  common,  and  the  contained 


angles  equal,  the  bases  LP,  MP,  NP  are  (I.  3)  all  equal.  Also 
(const  )  the  square  of  AB  is  equivalent  to  the  squares  of  LO, 
OP;  and  (I.  24,  cor.  1)  the  vsquare  of  LP  is  also  equivalent  to 
the  squares  of  LO,  OP,  because  LOP  is  a  right  angle.  There- 
fore (I.  ax.  1)  the  square  of  AB  is  equal  to  the  square  of  PL, 
and  (I.  23,  cor.  3)  AB  to  PL ;  and  hence  all  the  straight  lines 
LP,  IMP,  NP,  liA,  BC,  ED,  etc.,  are  equal.  Then,  in^lhe  tri- 
angles LPM,  ABC,  the  sides  AB,  BC  are  equal  to  LP,  PM, 


BOOK   VI.] 


EUCLID   AND   LEGENDRE. 


155 


eacli  to  each  ;  and  (const.)  AC  is  equal  to  LM ;  therefore  (I.  4) 
the  angles  AI>C,  LPM  are  equal ;  and  it  would  be  shown  in  a 
similar  manner,  that  the  angle  E  is  equal  to  ]\IPN,  and  H  to 
NPL.  The  solid  angle  at  P,  therefore,  being  contained  by- 
three  plane  angles,  which  are  equal  to  the  three  given  angles, 
B,  E,  H,  each  to  each,  is  such  as  was  required. 

Piiop.  XTI. — Theor. — A  plane  cutting  a  solid.,  and  parallel 
to  two  of  its  opposite  planes.,  divides  the  whole  into  two  solids^ 
the  base  of  one  of  which  is  to  the  base  of  the  other  as  the  one 
solid  is  to  the  other. 

Let  the  solid  BC  be  cut  by  the  plane  GF  which  is  parallel  to 
the  opposite  planes  BY  and  IIC,  and  divides  the  whole  into 
two  s(  lids,  BFand  GC;  as  the  base  of  the  first  is  to  the  base  of 
the  second,  so  is  BF  to  GC. 


N         M         H         G         B 


./LTP 


/■ 


V 


LJ-L 


V 


t 


Q 


K 


Produce  YC  both  ways,  and  take  on  one  side  any  number  of 
straight  lines,  YK  and  KL,  each  equal  to  YF,  and  complete 
parallelograms  similar  and  equal  to  BY.  Then,  because  BY, 
YK,  and  KL  are  all  equal,  the  parallelograms  on  them  are  also 
equal  (I.  15,  cor.  5);  and  for  same  reason  the  parallelograms 
on  the  other  side  of  YC,  on  the  straight  lines  CQ,  QS,  each 
equal  to  FC,  are  also  equal ;  therefore  three  ])lanes  of  the  solid 
XK  are  equal  and  similar  to  three  planes  of  KB,  as  also  to 
three  planes  of  YG.  But  (VL  9,  cor.  5)  the  three  planes  oppo- 
site to  these  three  are  equal  and  similar  to  them  hi  the  several 
solids ;  and  none  of  their  solid  angles  are  contained  bv  more 
than  three  plane  angles;  therefore  (VI.  9,  cor.  4)  the  solids 
XK,  KB,  and  YC  are  equal.  For  the  same  reason,  FH,  CM 
and  QN  are  also  equal.  Therefore  whatever  multiple  the  base 
LF  is  of  YF,  the  same  multiple  is  the  solid  LG  of  YG.  For 
same  reason,  whatever  multiple  the  base  FS  is  of  FC,  the  same 


156  THE   ELEMENTS   OF  [BOOK   TI. 

multiple  is  the  solid  FT  of  FH.  And  if  the  base  LF  be  equal 
to  SF,  the  solid  LG  is  equal  (VI.  9,  cor.  4)  to  FT ;  if  greater, 
greater;  and  if  less,  less.  Therefore  (IV.  def  5)  as  the  base 
YP^  is  to  the  base  FC,  so  is  the  solid  BF  to  the  solid  GC ; 
wherefore,  a  plane,  etc. 

Prop.  XTII. — Prob. — At  a  given  point  in  a  given  straight 
line,  to  make  a  solid  angle  equal  to  a  given  solid  angle  conr 
tained  by  three  plane  angles. 

Let  A  be  a  given  point  in  a  given  straight  line  AB,  and  D  a 
given  solid  angle  contained  by  the  three  plane  angles  EDO, 
EDF,  F'DC;  it  is  required  to  make  at  A  in  the  straight  line 
AB  a  solid  an  le  equal  to  the  solid  angle  D. 

In  DF  take  any  point  F,  from  which  draw  (VI.  5)  FG  per- 
pendicular to  the  plane  EDO,  meeting  that  plane  in  G ;  join 
DG,  and  (I.  13)  make  the  angle  BAL  equal  to  EDO,  and  in 
the  plane  BAL  make  the  angle  BAK  equal  to  EDG ;  then 
make  AK  equal  to  DG,  and  (VI.  6)  draw  KH  perpendicular  to 
the  plane  BAL,  and  equal  to  GF,  and  join  AH.  Then  the 
solid  angle  at  A,  which  is  contained  by  the  plane  angles  BAL, 
BAH,  HAL,  is  equal  to  the  given  solid  angle  at  D. 

Take  AB  DE  equal  to  one  another;  and  join  IIB,  KB,  FE, 
GE;  and  (VI.  def  1)  because  FG  is  perpendicular  to  the  plane 
EDO,  FGD,  F'GE  are  right  angles.  For  the  same  reason, 
HKA,  HKB  are  right  angles;  and  because  KA,  AB  are  equal 
to  GD,  DE,  each  to  each,  and  contain  equal  angles,  BK  is 
equal  (I.  3)  to  EG ;  also  KH  is  equal  to  GF,  and  HKB,  FGE 
are  right  angles ;  therefore  HB  is  equal  to  FE.  Again :  be- 
cause AK,  KH  are  equal  to  DG,  GF,  and  contain  right  angles. 


AH  19,  equal  to  DF;  also  AB  is  equal  to  DE,  and  IIB  to  FE; 
therefore  (L  4)  the  angles  BAH,  EDF  are  equal.     Again :  since 


BOOK  VI.]         EUCLID  AND  LEGENDRE.  157 

(const.)  the  angle  BAL  is  equal  to  EDC,  and  BAK  to  EDG, 
tlie  remaining  angles  KAL,  GDC  are  (I.  ax.  3)  eqnal  to  one 
another;  and,  by  taking  AL  and  DC  eqnal,  and  joining  LH, 
LK,  CF,  CG,  it  would  be  proved,  as  in  the  foregoing  part,  that 
the  angle  HAL  is  equal  (I.  4)  to  FDC.  Therefore,  because  tlie 
three  plane  angles  BAL,  BAII,  HAL,  which  contain  the  solid 
angle  at  A,  are  equal  to  the  three  EDC,  EDF,  FDC,  which 
contain  the  solid  angle  at  D,  each  to  each,  and  are  situated  in 
the  same  order,  the  solid  angle  at  A  is  equal  (VL  9,  cor.  3)  to 
the  solid  angle  at  D.  Therefore,  what  was  required  has  beeu 
done. 

Prop.  XIV. — Theor. — If  a  parallclopiped  he  cut  by  a  plane 
passing  through  the  diagonals  of  two  of  the  opposite  planes^ 
it  is  bisected  by  that  plane. 

Let  AB  be  a  parallclopiped,  and  DE,  CF  the  diagonals  of 
the  opposite  parallelograms  AH,  GB,  viz.,  those  which  join  the 
equal  angles  in  each.  Then  (VI.  3,  cor.  2)  CD,  P^E  are  paral- 
lels, because  each  of  them  is  parallel  to  GA  ;  wherefore  (VL  3, 
cor,  1)  the  diagonals  CF,  DE  are  in  the  plane  in  which  the  par- 
allels are,  and  (VL  8)  are  themselves  parallels.  Again:  be- 
cause (L  15,  cor.  1)  the  triangle  CGF  is  equal  to  CBF,  and 
DAE  to  DHE;  and  that  (VL  12)  the  parallelogram  CA  is 
equal  and  similar  to  the  opposite  one  BE;  and 
GE  to  CH ;  therefore  the  prism  contained  by 
the  two  triangles  CGF,  DAE,  and  the  three 
parallelograms  CA,  GE,  EC,  is  equal  (VL  9, 
cor.  4)  to  the  prism  contained  by  the  two  tri- 
angles CBF,  DHE,  and  the  three  parallelo- 
grams BE,  CH,  EC ;  because  they  are  contained  by  the  same 
number  of  equal  and  similar  planes,  alike  situated,  and  none  of 
their  solid  angles  are  contained  by  more  than  three  plane  an- 
gles.    Therefore,  if  a  parallclopiped,  etc. 

Seho.  The  insisting  litres  of  a  parallclopiped  are  the  sides  of 
the  parallelograms  between  the  base  and  the  opposite  plane. 

Cor.  In  a  parallclopiped,  if  the  sides  of  two  of  the  opposite 
planes  be  each  bisected,  the  common  section  of  the  planes  pass- 
ing through  the  points  of  division,  and  any  diagonal  of  the  par- 
allelepiped bisect  each  other. 


158 


THE   ELEMENTS   OF 


[book  VI. 


Pkop.  XV. — Theor. — ParaUelopipeds  upon  the  same  base^ 
and  of  the  same  altitude,  the  insisting  lines  of  which  are  ter- 
minated in  the  same  straight  lines  of  the  />^t</ie  opposite  to  the 
base,  are  equal  to  one  another. 

Let  the  parallelopipeds  All,  AK  (2fl  fig.)  be  upon  tlie  same 

base  AB,  and  of  tlie  same  altitude;  and  let  their  insisting  lines 

AF,  AG,  LM,  LN  be  terminated  in  the  same  straight  line  F'N, 

and  CD,  CE,  BH,  BK  in  the  same  DK ;  AH  is  eqlial  to  AK. 

First,  let  the  parallelograms  DG,  HN,  whicli  are  opposite  to 

the  base  AB,  have  a  comtnon  side 
HG.  Then  because  AH  is  cut  by  the 
plane  AGHC  passing  through  the  di- 
agonals AG,  CH  of  the  opposite  planes 
ALGF,  CBHD,  All  is  bisected  (VI. 
14)  by  the  plane  AGHC.  For  the 
A  L  same  reason,  AK  is  bisected  by  tlie 

plane  LGHB  through  the  diagonals 
LG,  BH.  Therefore  the  solids  AH,  AK  are  equal,  each  of 
them  being  double  of  the  prism  contained  between  the  trian- 
gles ALG^  CBH. 

But  let  the  parallelograms  DM,  EX,  opposite  to  the  base, 
have  no  common  side.  Then,  because  CH,  CK  are  parallelo- 
grams, CB  is  equal  (I.  15,  cor.  1)  to  each  of  the  ojjposite  sides 
DH,  EK;  wherefore  DH  is  equal  to  EK.  P^'rom  1)K  take  sep- 
arately EK,  DH;  then  DE  is  equal  to 
HK  ;  wherefore,  also  (I.  15,  cor.  5), 
the  ti'iangles  CDE,  BHK  are  equal ; 
and  (I.  15,  cor.  5)  the  parallelogram 
DG  is  equal  to  HN.  For  the  same 
reason,  the  triangle  AFG  is  equal  to 
LMN;  and  (VI.  12)  the  parallelo- 
gram CF  is  equal  to  BM,  and  CG  to  BN ;  for  they  are  oppo- 
site. Therefore  (VI.  9,  cor.  4)  the  prism  which  is  contained  by 
the  two  triangles  AFG,  CDE,  and  the  three  jiarallelograms 
AD,  DG,  GC,  is  equal  to  the  prism  contained  by  the  two  tri- 
angles LMN,  BHK,  and  the  three  parallelograms  BM,  MK,  KL. 
If,  therefore,  the  prism  LMNBHK  be  taken  from  the  solid  of 
which  the  base  is  the  parallelogram  AB,  and  in  which  FDKN 
is  the  one  opposite  to  it ;  and  if  from  the  same  solid  there  be 


BOOK  VI.] 


EUCLID  AND  LEGENDRE. 


150 


taken  the  prism  AFGCDE,  the  remaining  solids  AH,  AK  are 
equal.     Therefore,  parallelopipeds,  etc. 

Cor.  1.  Also  parallelopipeds  upon  the  same  base  and  of  the 
same  altitude,  the  insisting  lines  of  which  are  not  terminated  in 
the  same  straight  lines  in  the  plane  opposite  to  the  base,  are 
equal  to  one  another. 

Cor.  2.  Hence  parallelopipeds  which  are  upon  equal  bases, 
and  of  the  same  altitude,  are  equal  to  one  another  (I.  15, 
cor.  5). 

Co::  3.  Parallelopipeds  which  have  the  same  altitude,  are  to 
one  another  as  their  bases  (VI.  12). 

Cor.  4.  If  thei-e  be  two  triangular  prisms  of  the  same  alti- 
tude, the  base  of  one  of  which  is  a  parallelogram,  and  that  of 
the  other  a  triangle ;  if  the  parallelogram  be  double  of  the  tri- 
angle, the  prisms  are  equal  (I.  15,  cor.  4). 


o 


M 


-o4 


V. 


y\ 


jy 


Vti 


N 


Prop.  XVI. — Theor. — Similar  solids  are  one  to  another  in 
the  tnplicate  ratio  of  their  homologous  sides. 

Let  AB,  CD  be  similar  parallelopipeds,  and  the  side  AE 
homologous  to  CF ;  AB  has  to  CD  the  triplicate  ratio  of  that 
which  AE  has  to  CF. 

Produce  AE,  GE,  HE,  and  in  these  produced  take  EK  equal 
to  CF,  EL  equal  to  FN,  and 
EM  equal  to  FR ;  and  com- 
plete the  parallelogram  KL 
and  the  solid  KO.  Because 
IvE,  EL  are  equal  to  CF,  FN, 
and  the  angle  KEL  equal  to 
the  angle  CFN,  since  it  is 
equal  to  the  angle  AEG, 
which  is  equal  to  CFN,  be- 
cause the  solids  AB,  CD  are 

similar;  therefore  the  parallelogram  KL  is  similar  and  equal  to 
CN.  For  the  same  reason,  the  parallelogram  MK  is  similar 
aii<l  equal  to  CR,  and  also  OE  to  FD.  Therefore  three  paral- 
lelograms of  the  solid  KO  are  equal  and  similar  to  three  paral- 
lelograms of  the  solid  CD;  and  (VI.  12)  the  three  opposite 
ones  in  each  solid  are  equal  and  similar  to  these.  Therefore 
(VI.  9,  cor.  4)  the  solid  KO  is  equal  and  similar  to  CD.     Com- 


100 


THE    FXKMENT8    OF 


[book  VI. 


N 


"sC 


Jj 


R 


plete  the  paralIeloG:rani  GK,  and  the  solids  EX,  LP  upon  the 
bases  GK,  KL.  so  that  EH  may  be  an  insisting  line  in  each  of 
them ;  and  thus  they  are  of  the  same  altitude  with  the  solid 

>^B.  Then,  because  the  sol- 
ids AB,  CD  are  similar  (VI. 
def.  8,  and  alternately),  as 
AE  is  to  CF,  so  is  EG  to  FN, 
and  so  is  EH  to  FR ;  and  FG 
is  equal  to  EK,  and  FN  to 
EL,  and  FR  to  EM  ;  there- 
fore, as  AE  to  EK,  so  is  p]G 
to  EL ;  and  so  is  HE  to  EM. 
But  (V.  1)  as  AE  to  EK,  so 
is  the  parallelogram  AG  to  GK ;  and  as  GE  to  EL,  so  is  GK 
to  KL ;  and  as  HE  to  EM,  so  is  BE  to  KM ;  therefore  (IV.  7) 
as  the  parallelogram  AG  to  GK,  so  is  GK  to  KL,  and  PE  to 
KM.  But  (VI.  13)  as  AG  to  GK,  so  is  the  solid  AB  to  EX ; 
and  as  GK  to  KL,  so  is  the  solid  EX  to  PL ;  and  as  PE  to 
KM,  so  is  the  solid  PL  to  KG;  and  therefore  (IV,  7)  as  the 
solid  AB  to  EX,  so  is  EX  to  PL,  and  PL  to  KG.  But  if  four 
magnitudes  be  continual  proportionals,  the  first  is  said  to  have 
to  the  fourth  the  trijtlicate  ratio  of  that  which  it  has  to  the  sec- 
ond; therefore  the  solid  AB  has  to  KG  the  triplicate  ratio  of 
that  which  AB  has  to  EX.  But  as  AB  is  to  EX,  so  is  the  par- 
allelogram AG  to  GK,  and  the  straight  line  AE  to  EK. 
Wherefore  the  solid  AB  has  to  the  solid  KG  the  triplicate  ratio 
of  that  which  AE  has  to  EK;  and  the  solid  KG  is  equal  to  CD, 
and  the  straight  line  EK  to  CF.  Tliei-efore  the  solid  AB  has 
to  CD  the  ti-i]tlicate  ratio  of  that  which  the  side  AE  has  to  the 
homologous  side  CF. 

Since  (VI-  15)  CNF  is  half  of  the  base  of  CD,  CRF  half  of  CR, 
and  NRF  half  of  NR,  which  planes  form  the  triangular  pyramid 
CNF,  K  ;  and  since  the  triangular  pyramid  AGE,  II  is  formed  ia 
a  similai"  manner,  and  the  parallelo])iped  AB  has  to  the  paral- 
lcl()pi|)C(l  CI)  the  triplicate  ratio  of  that  which  the  side  AE 
has  to  tlie  side  CF  ;  hence  (IV,  ax.  1)  the  triangular  ])yramid 
CNF,  R  has  to  the  triangular  j)yrainid  AGE,  II  the  triplicate 
ratio  of  tliat  which  AE  has  to  CF.  And  (L  20,  cor.  1)  all 
polygons  can  be  divided  into  triangles;  therefore  (V.  14)  solids 


BOOK  VI.]         EUCLID  AND  LEGENDRE.  131 

on  similar  bases  liave  to  one  another  tlie  triplicate  ratio  of 
lioni<)'o<r()iis  >i(les.     Wherefore,  siniihir  solids  aie,  etc. 

Cor.  1.  Hence,  similar  solids  of  the  same  or  equal  bases  are 
to  one  another  as  their  altitudes — and,  conversch',  those  of  the 
same  ov  equal  altitudes  are  to  one  another  as  their  bases. 

Cor.  2.  Hence,  also,  the  bases  and  altitudes  of  equivalent 
polids  are  reciprocally  proportional;  and  conversely,  solids 
lia villi?  their  bases  and  altitudes  reciprocally  proportional,  are 
equivalent. 

Cor.  3.  Hence,  cones  and  cylinders  upon  equal  bases  are  a3 
their  altitudes,  and  their  bases  and  altitudes  are  reciprocally 
proportional  when  the  cones  and  cylinders  are  equivalent. 
And  similar  cones  and  cylinders  have  to  one  another  the  tripli- 
cate ratio  of  that  which  the  diameters  of  their  bases  have  ;  and 
gpheres  have  the  triplicate  ratio  of  their  diameters  (V.  14  and 

VI.  1(3). 

Cor.  4.  From  this  it  is  manifest,  that  if  four  straight  lines  be 
continual  proportionals,  as  the  first  is  to  the  fourth,  so  is  the 
parallelopiped  described  from  the  first  to  the  similar  solid  simi- 
larly described  from  the  second  ;  because  the  first  strais^ht  line 
has  to  the  fourth  the  triplicate  ratio  of  that  which  it  has  to  the 
second. 

Cor.  5.  Parallelopipeds  contained  by  parallelograms  equi- 
angular to  one  another,  each  to  each,  that  is,  of  which  the  solid 
angles  are  equal,  each  to  each,  have  to  one  another  the  ratio 
which  is  the  same  with  the  ratio  compounded  of  the  ratios  of 
their  sides  (V.  14,  cor.  7). 

Cor.  6.  The  bases  and  altitudes  of  equivalent  parallelopipeds 
are  reciprocally  proportional;  and  (2)  if  the  bases  and  altitudes 
be  reciprocally  proportional,  the  pai'alleloj)ipeds  are  equivalent. 

Prop.  XVH. — Tiieor. — Every  pyramid  is  one  third  the 
prism  of  the  same  b-fse  and  altitude,  and  every  co?ie  is  one 
third  of  the  cylinder  with  the  same  base  and  altitude.,  or  every 
pyramidal  solid  is  one  third  the  solid  of  the  same  base  and 
altitude. 

Let  there  be  a  prism  of  which  the  bases  are  the  triangles 
ABC,  DEF ;  the  prism  may  be  divided  into  three  equal  tri- 
angular pyrajnids. 
11 


162 


THL    ELEMENTS    OF 


[book  VL- 


Join  BD,  EC,  CD;  and  because  ABED  is  a  parallelogram, 
and  BD  its  diagonal,  the  triangles  ABD,  EBD  are  (I.  15,  cor. 
1)  equal;  therefore  (VI.  IC,  cor.  3)  the  pyramid  of  which  the 
base  is  ABD,  and   vertex  C,  is  equal  to  tho 
pyramid  of  which  the  base  is  EBD  and  vertex 
C.     But  EBC  may  be  taken   as  the  base  of 
this  pyramid,  and  D  as  its  veitex.    It  is  there- 
fore equal  (VI.  16,  cor.  3)  to  the  pyramid  of 
which  ECF  is  the  base,  and  D  the  vertex;  for 
they  l)ave  the  same  altitude,  and  (I.  15,  cor.  1) 
equal  bases  ECF,  ECB;  and  it  has  been  al- 
ready  proved   to    be   equal    to   the   pyramid 
ABDC.     Therefore   the   prism   A13CDEF  is 
divided  into  three  equal  pyramids  having  tri- 
angular bases,  viz.,  into  the  pyramids  ABDC,  EBDC,  ECFD. 
Therefore,  every  ti'iangular  piism,  etc. 

Now  every  polygon  can  be  divided  into  triangles  (I.  20,  cor. 
1)  ;  hence  every  j)risni  with  a  polygonal  base  can  be  divided 
into  i)risms  havinij  trianirular  bases :  and  as  each  of  these  tii- 
angu.lar  prisms  can  be  divided  into  three  equivalent  triangular 
pyramids,  therefjre  every  pyramid  is  one  third  the  prism  of 
Bame  base  and  altitude.  As  it  has  been  shown  (V.  14)  that 
nil  surfaces  are  to  one  another  in  the  duplicate  ratio  of  their 
liomologous  sides,  and  (VI.  IC)  all  solids  are  to  one  another  in 
the  trij)licate  ratio  of  their  homologous  sides,  it  follows  that  all 
Bolids  of  similar  bases  and  altitudes  have  the  same  propoition 
to  one  another  (VI.  16,  cor.  3);  hence,  cones  having  similar 
bases  and  altitudes  to  cylinders,  have  the  same 
proportion  to  those  cylinders  which  pyramids 
have  to  prisms  of  similar  bases  and  altitudes; 
therefore  the  cones  are  one  thiid  the  cylinders, 
as  it  is  evident  that  the  section  of  the  cone  and 
cylinder  is  similar  to  the  section  of  the  pyramid 
and  prism  of  whatever  regular  and  similar  base. 
For  ABC  can  be  the  section  of  a  cone  and  of  a 
pyramid  of  any  regular  base,  and  ABED  can  be 
the  section  of  a  prism  of  any  similar  base,  and 
pcction  of  a  cylinder  of  similar  base  with  cone ;  therefore 
the  same  proportion  which  regulates  the  respective  magnitudes 


BOOK  VI,] 


EUCLID  AND  LEGENDRE. 


163 


of  tlie  pyramid  and  pi-ism,  also  regulates  the  respective  mag- 
nitudes of  cone  and  cylinder — as  all  surfaces  are  (V.  14)  to 
one  another  in  the  duplicate  ratio  of  their  homologous  sides, 
and  (VI,  16)  all  solids  are  to  one  another  in  the  triplicate  ratio 
of  their  homologous  sides. 

Cor.  1,  p]very  sphere  is  two  thirds  of  its  circumscribing  cyl- 
inder. Let  ABCD  be  a  cylinder  circumscribing  the  sphere 
EFGH  ;  then  will  the  sphere  EFGH  be  two  thirds  of  the  cylin- 
der ABCD,  For,  let  the  plane  AC  be  a  section  of  the  sphere 
and  cylinder  through  the  center  J,  Join  AJ,  B.T.  Also,  let 
FJII  be  paiallel  to  AD  or  BC,  and  OKL  be  parallel  to  AB  or 
DC,  the  base  of  the  cylinder,  the  latter  line,  KL,  meeting  BJ 
in  M,  and  the  circular  section  of  the  sphere  in  N. 

Then,  if  the  whole  plane   HFBC  be  conceived  to   revolve 
about  the  line  IIF  as  an  axis,  the  squai'e 
FG   will   describe  a  cylinder,  AG  ;  the 
quadrant  JFG  will    describe   a    hemis- 
phere, EFG;  and  the  triangle  JFB  will 

describe  a  cone,  JAB.    Also,  in  the  rota-      El ^^ |o 

tion,  the  three  lines  or  parts  KL,  KN, 
KM,  as  radii,  will  describe  correspond- 
ing circular  sections  of  those  solids,  viz.,        D  u  c 
KI^,  a  section  of  the  cylinder;  KN,  a 

«;ection  of  the  sphere;  and  KM,  a  section  of  the  cone.  Now, 
FB  being  equal  to  FJ  or  JG,  and  KL  parallel  to  FB,  then  by 
eimilar  triangles  JK  is  equal  to  KM.  And  since  in  the  right- 
angled  triangle  JKN,  JN-OJK--|-KN-  (L  24,  cor.  1),  and  be- 
cause KL  is  equal  to  the  radius  JG  or  JN,  and  KM  is  equal  to 
JK,  therefore  KL-<;>KM"-|-KN-;  and  because  circles  areas 
the  squares  of  their  diameters  or  the  squares  of  their  radii, 
therefore  (V.  14,  cor.  2)  circle  of  KL  is  equivalent  to  circles  of 
KM  and  KN,  or  section  of  cylinder  is  equivalent  to  both  cor- 
responding sections  of  sphere  and  cone.  And  as  this  will 
always  be,  it  follows  that  the  cylinder  EB,  which  is  all  the 
fonner  sections,  is  equivalent  to  the  hemisphere  EFG  and  cone 
JAB,  which  are  all  the  latter  sections.  But  JAB  is  one  third 
of  the  cylinder  EB  (VI.  1 7) ;  therefore  (I.  ax.  3)  the  hemisphere 
EFG  is  two  thirds  of  the  cylinder  EB. 

Cor.  2.  If  the  parallelogram  BEGC  be  revolved  around  the 


K^ 

mX 

'  \ 

/^ 

J 

V 

y 

16i 


THE   ELEMENTS   OF 


[book  VI. 


fixed  axis  BC,  it  will  generate  a  cylincler  (VI.  def.  24) ;  the 
semicircle  BNC  will  generate  a  sphere  (VI.  def.  17)  ;  and  the 
triangle  BGC  will  generate  a  cone  (VI.  def.  21).  The  cone 
Avill  be  one  third  the  cylinder  (VI.  17),  and  the  sphere  will  bo 
two  thirds  the  same  cylinder  (VI.  17,  cor  1). 

The  triangle  BOP  having  one  half  altitnde  and  one  half  base 
of  the  triangle  BGC,  will  generate  a  cone 
one  eighvh  of  the  cone  generated  by  the 
triangle  BGC  (VI.  16,  cor.  3)  ;  hence,  one 
twelfth  of  the  cylinder  generated  by  the 
square  BENP;and  the  cone  generated  by 
the  triangle  BNP  is  one  half  cone  gener- 
ated by  the  triangle  BGC  (VI.  16,  cor.  1) ; 
hence,  four  times  cone  generated  by  the 
triangle  BOP.  And  the  hemisphere  gen- 
erated by  the  quadrant  BNP  is  two  thirds 
cylinder  generated  by  the  square  BENP 
VI.  17,  cor.  1),  or  eight  times  cone  gen- 
erated by  the  triangle  BOP. 

Let  the  triangle  BSN  be  described  on  BN,  equal  to  the  tri- 
angle BON  (I.  23,and  15,  cor.  4).  Then  the  trapezium  BSNP 
will  generate  a  solid  equivalent  to  the  sum  of  a  cylinder  one 
half  cylinder  generated  by  the  square  BENP,  and  a  cone  one 
sixth  of  the  same  cylinder,  or  eight  times  the  cone  generated 
by  the  triangle  BOP,  making  a  solid  equivalent  to  the  hemis- 
phere generated  by  the  quadrant  BNP  on  the  same  radius  PN" 
and  same  altitude  BP.  But  the  triangle  BNP  is  common  to 
both  the  trapezium  B>NP  and  the  quadiant  BNP,  and  gener- 
ates in  each  case  the  solid  equivalent  to  four  times  cone  gener- 
ated by  the  tiiangle  BOP ;  therefore  the  segment  BN  and  the 
triangle  BSN  generate  an  equivalence  of  solid,  or  four  times 
cone  generated  by  the  triangle  BOP;  consequently  the  seg- 
ment BN  and  the  triangle  BSN  are  equivalent  (I.  ax.  1). 

Again :  the  triangle  BNP  generates  a  cone  one  third  the 
cylinder  generated  by  the  square  BENP  (VI.  17),  and  the 
quadrant  BNP  generates  a  heniisphere  two  thiids  of  the  same 
cylinder  (VI.  17,  cor.  1).  The  triangle  BNP  is  one  half  the 
square  BlilNP. 

Now,  the  trapezium  BSNP,  equivalent  to  three  fourths  of  the 


BOOK  VI.]  EUCLID  AND  LEGENDRE.  165 

gquiire,  on  same  raflius  aiul  altitude  as  the  square,  generates 
a  solid  two  thirds  of  the  solid  generated  by  the  sqnuiv,  and  the 
quadrant  BNP  witli  same  radius  and  altitude  as  the  square 
BENP  generates  an  equivalent  solid  with  the  trapezium 
BSNP.  That  an  equivalence  of  surfaces  ujion  the  same  radius 
•will  generate  an  equivalence  of  solids  can  be  illustrated  by- 
taking  a  trapezium  greater  than  the  trapezium  BSNP,  having 
tlie  same  radius.  It  can  easily  be  shown  that  the  greater  trape- 
zium generates  a  greater  solid  than  the  less  trapezium,  and  in 
a  similar  manner  it  can  be  shown  that  a  less  trapezium  than  the 
trapezium  BSNP  generates  a  less  solid  ;  hence,  very  evidently, 
when  a  greater  surface  upon  same  radius  generates  a  greater 
solid,  and  a  less  surface  generates  a  less  solid,  equivalent 
surfaces  must  generate  equivalent  solids  on  the  same  radius; 
and,  conversely,  when  we  have  equivalent  solids  generated 
npon  the  same  radius,  the  generatirig  surfaces  are  equivalent ; 
therefore  (I.  ax.  l)  the  quadrant  BNP  is  three  fourths  of  the 
square  BENP,  or  the  semicircle  BNC  is  three  fourths  of  the 
parallelogram  BEGC,  or  any  circle  is  three  fourths  of  the  cir- 
cumscribing square,  or  Three  Times  Square  of  Badius. 
Hence,  we  have  a  geometrical  confirmation  of  the  mechanical 
construction  in  scholium  to  twenty-fifth  proposition  of  book 
fifth. 

<  TIIER'WISR  : 

Tlie  triangle  BGC  generates  a  cone  one  third  (VI.  17)  of 
the  cylinder  generated  by  tlie  rectangle  BEGC,  and  the  semi- 
cii'cle  BXC  genei'ates  a  sphere  two  thirds  ( VI.  1  7,  cor.  1 )  of  the 
same  cylinder ;  the  s]  here  is  the  mean  between  the  cone  and 
cylinder;  therefore  the  semicircle  is  evidently  the  meanhe- 
tween  the  triangle  BGC  and  the  rectangle  BEGC,  or  three 
fourths  of  the  rectangle  BEGC  ;  or  any  circle  is  three  fourths 
square  of  its  diameter,  or  three  times  square  of  its  radius. 

OTHERWISE  : 

Circles  are  to  one  another  as  the  squares  described  on  their 
diameters  (V.  14)  ;  consequently  squares  are  to  one  another  as 
the  circles  described  on  their  sides;  therefore  there  is  an 
equality  of  proportion  (V.  24)  ;  hence,  we  derived  the  arith* 
metical  proportion : 


166  THE   ELEMENTS   OF  [boOK   YI. 

Rectangle  BEGC,  semicircle  BNC,  triangle  BGC. 

The  sum  of  extremes  is  equivalent  to  twice  the  mean ;  there- 
fore we  have — 

Rectangle  BEGC -f- triangle  BGCo2  semicircle  BNC;  or, 
semicircle  BNCOi  rectangle  BEGC  +  i  triangle  BGC. 

Also,  the  difference  between  first  and  second  terms  of  an 
arithmetical  proportion  is  the  same  as  the  difference  between 
the  second  and  third  terms,  as  the  diffei'ence  between  third  and 
fourth  terms,  and  so  on  ;  hence  we  have — 

Rectangle  BEGC— semicircle  BNCOsemicircle  BNC -tri- 
angle BGC  ;  therefore  we  get  segment  BXOtriangle  BSN, 
or  one  fourth  square  BENP;  consequently,  eircle  =0  three 
fourth  square  of  diameter,  or  tliree  times  square  of  radius. 

Cor.  3.  Archimedes  discovered  the  proi)ortion  1,  2,  3  be- 
tween the  cone,  sjihere,  and  cylinder  of  similar  dimensions;  but 
from  the  previous  corollary  we  obtain  the  proportion  1,  2,  3, 
4  for  the  cone,  sphere,  cylinder,  and  cube  of  similar  diuiensions; 
because  the  cube  is  eight  times  cube  of  radius  of  th^  sjyhere 
(Vr.  10) ;  the  cylinder  is  six  times  cube  of  radius  of  the  sphere 
(VI.  17,  cor,  2)  ;  the  sphere  infour  times  cube  of  radius  (f  the 
sphere  (VI.  17,  cors.  1  and  2)  ;  and  the  cone  is  twice  ctibe  of 
radius  of  the  sphere  (VI,  17,  cors,  1  and  2).  Hence  the  sphere 
is  the  mean  proportional  of  the  cone,  and  the  cube  circum- 
scribing the  si)here,  or  one  half  the  circumscribiug  cube; 
therefore  the  surface  of  the  sphere  is  fur  times  the  area  of 
one  of  its  great  circles,  or  two  thirds  the  surface  of  the  circum- 
scribing cylinder.  Hence,  there  is  the  identical  proportion  be- 
tween the  surfaces  of  the  sphere  and  cylinder  as  there  is 
between  \}\q  solidities  of  the  s[ihere  and  cylinder. 

Scho.  1.  Therefore  the  second  corollary  gives  the  solution 
to  the  long  mooted  and  much  vexed  question  of  the  Quadra- 
ture of  the  Circle,  showing  that  the  perplexity  of  it  arose  from 
the  uii geometrical  sitjyposition  (V.  25,  scho.)  that  "the  circle  is 
a  regular  polygon  of  an  infinite  inind)er  of  sides."  Hence  it  is 
evident  that  all  conclusions  derived  fi'oni  a  fdlacious  su])])0- 
sition  will  give  ])erplexity  so  long  as  the  snpposilion  is  main- 
tained, and  must  necessarily  involve  conti'adiction*^  to  the  rigor 
of  geometrical  reasoning.  And  wlicii  demonstrations  are  con- 
ducted consistently  with  established  definitions,  axioms,  and 


BOOK    VI.]  EUCLID   AND   LKGKNDRE.  16T 

propositions,  all  conclusions  derived  from  tlieni  are  unimpeach- 
able, and  arc  valuable  to  a  system  of  scientific  truths. 

iicho.  2.  Geometry,  like  all  other  sciences,  is  based  upon 
cvviiuuj'undamencal  i)rinciples,  and  a  close  examination  of  tliia 
science  reveals  the  fact  that,  throughout  its  whole  extent  and 
its  vai'ious  applications,  tlie  principle  that  the  siraljht  line  is 
the  shortest  line  bcttoeen  tv:o  fjiven  points  is  the  fundamental 
])rinciple  of  the  science  ;  by  this  principle  the  dimensions  of 
magnitudes  are  determined,  distances  of  objects  made  known, 
and  other  useful  and  practical  results  ascertained.  iSince  this 
jjrinciple  is  so  important,  it  would  be  interesting  to  inquire  the 
reason  why  it  has  such  manifest  usefulness.  The  angle  is  a 
magnitude  contained  l)y  the  intersection  of  two  straifjht  lii.es^ 
and  the  polygon  is  another  magnitude  bounded  by  three  or 
moYG  straight  lines  ^  lience  we  see  how  intimate  the  connec- 
tion between  the  straight  line  and  the  angle  and  polygon ; 
therefore  we  find  that  the  functions  of  the  angle  are  straight 
lines,  such  as  sines,  co-sines,  tangents,  etc. ;  and  the  properties 
of  the  polygon  ai"e  defined  by  straigld  lines,  such  as  its  jjerime- 
ter  and  apothem ;  therefore  in  all  rectilineal  magnitudes  we 
discover  a  use  for  the  straight  line  above  all  other  lines,  and 
evidently  the  principle  of  the  straight  line  has  a  i)eculiar  force 
to  all  rectilineal  figures;  consequently  we  adopt  x}w  straight 
line  as  a  means  of  measure  for  all  rectilineal  magnitudes.  The 
adoption  of  this  m^eans  of  measure  constitutes  the  straight  line 
a  standard  by  which  all  measurements  of  rectilineal  magni- 
tudes are  compared.  Hence  very  naturally  there  is  a  consist- 
ency between  the  measurements  and  other  properties  of  recti- 
lineal magnitudes. 

Now,  when  w(^  examine  the  circle  or  portions  of  the  circle, 
as  the  segments,  sectors,  arcs,  etc.,  we  at  once  discover  a  nou' 
coincidence  between  the  curve  which  bounds  them  and  the 
straight  line  which  bounds  rectilineal  magnitudes ;  hence,  very 
evidently,  the  superficies  of  curvilmear  spaces  require  a  pecu- 
liar coni-ection  between  tiiem  and  the  bounding  curve,  as  there 
is  a  peculiar  connection  between  the  superficies  of  rectilineal 
spaces  and  the  bounding  straight  lines.  In  other  words,  since 
"vve  derive  the  measurements  and  other  properties  of  rectilineal 
magnitudes  from  the  principle  of  the  straight  line^  so  we  nmst 


V 


168 


THE   ELEMKNTS    OF 


[book  VI. 


determine  tlie  measurements  and  other  properties  of  curvilinear 
magnitudes  from  the  prini-iple  of  tlie  curve,  and  thus  we  per- 
ceive wh^,  when  we  endeavor  to  obtain  the  area  of  circle  by 
the  method  of  exhaustions,  using  the  straight  line  as  a  mean$ 
of  measure,  we  can  get  the  a2:>proxirnate  area  only,  and  ahy 
it  is  necessai-y  to  obtain  accurate  results  to  use  the  [trniciple  of 
the  curm.  Geometry,  in  its  pi'esent  state,  is  the  science  of  iho 
straight  line,  and  the  introduction  of  the  principle  of  t!ie  curve 
into  geometrical  consideration  would  usher  in  a  distinct  science^ 
but  eminently  useful  in  solving  problems  of  cui'vilincar  spaces 
and  boundaries  which  were  before  unsolved,  inasmuch  as  the 
approximate  results  only  were  given  for  them. 

The  method  of  exhaustions  is  applicable  to  rectilineal  magni- 
tudes, and  its  results  are  consistent  with  the  ])rinciple  of  the 
Btraight  line,  because  the  straight  line  is  adopted  as  a  means 
and  standard  oi'  measurement;  but  since  the  straight  line  and 
curve  do  not  coincide,  the  principle  and  propeities  of  the 
straight  line  are  not  applicable  to  curvilinear  spaces  or  bound 
aries ;  hence,  what  is  true  in  one  case,  becomes  absurd  in  the 
other. 


Prop.  XVIIT. — Tiieou.  —  77te  sections  of  a  solid  I y  parallel 
planes  are  similar  figures. 

Let  the  prism  JMN  be  cut  by  the  two  parallel  planes  AD, 
FK  ;  their  sections  with  it  are  similar  figures. 

For  (VI.  7,  cor.)  the  sections  have  parallel  sides  (T.  15,  cor. 
2).  The  figures  AD,  FK,  thei-efore,  have  their 
sides  similar,  each  to  each.  Their  several  an- 
gles are  also  (VI.  4)  equal  ;  for  they  ai"e  con- 
tained by  straight  lines  wliich  are  parallel ; 
and  therefore  the  figures  are  similar. 

Cor.  1.  A  section  of  a  prism  by  a  plane  par- 
allel to  the  base  is  equal  and  bimilar  to  the 
base. 

Scho.  1.  Since  (VI.  def.  24)  a  cylinder  is  de- 
scribed by  the  revolution  of  a  rectangle  about 
one  of  its  sides,  it  is   plain  that  any  straight 
line  in  the  rectangle  perpendicular  to  the  fixed 
line  will   describe  a  circle  parallel  to   the  base ;   and  hence 


BOOK  VI.]         EUCLID  AND  LEGENDEE.  169 

every  section  of  a  cylinder  by  a  plane  parallel  to  the  base  is  a 
circle  equal  to  the  base. 

Cor.  2.  The  section  of  a  pyramid  by  a  plane  parallel  to  its 
base  is  a  fiijure  similar  to  the  base. 

Scho.  2.  Since  (VI.  def.  21)  a  cone  is  described  by  the  revo- 
lution of  a  right-angled  triangle  about  one  of  its  legs,  it  is  plain 
that  any  straight  line  in  the  triangle  perpendicular  to  the  fixed 
leg  will  describe  a  circle  parallel  to  the  base;  and  the  radius 
of  that  circle  will  be  to  the  radius  of  the  base  as  the  altitude  of 
the  cone  cut  off  to  that  of  the  whole  cone. 

Cor.  3.  A  section  of  a  sphere  by  a  plane  is  a  circle. 

Since  the  radii  of  the  sphere  are  all  equal,  each  of  them  being 
equal  to  the  radius  of  the  describing  semicircle,  it  is  plain  that 
if  the  section  pass  through  the  center,  it  is  a  circle  of  the  same 
radius  as  the  sphere.  But  if  the  plane  do  not  pass  through  the 
center,  draw  (VI.  5)  a  perpendicular  to  it  from  the  center,  and 
draw  any  number  of  radii  of  the  sphere  to  the  intersection  of 
its  surface  with  the  plane.  These  radii,  which  are  equal,  are 
the  hypothenuses  of  right-angled  triangles,  which  have  the  per- 
pendicular from  the  center  as  a  common  leg;  and  therefore  (I. 
24,  cor.  2)  their  other  legs  are  all  equal ;  wherefore  the  section 
of  the  sphere  by  the  plane  is  a  circle,  the  center  of  which  is  the 
point  in  which  the  perpendicular  cuts  the  plane. 

8cho.  3.  All  the  sections  through  the  center  are  equal  to  one 
another,  and  are  greater  than  the  others.  The  former  are 
therefore  called  great  circles,  the  latter  small  or  less  circles. 

tScho.  4.  A  straight  line  drawn  through  the  center  of  a  circle 
of  the  sphere  perpendicular  to  its  plane  is  a  diameter  of  the 
sphere.  The  extremities  of  this  diameter  are  called  the  poles 
of  the  circle.  It  is  plain  (I.  24,  cor.  2)  that  chords  drawn  in 
the  sphere  from  either  pole  of  a  circle  to  the  circumfeience  are 
all  equal ;  and  therefore  (III.  16,  cor.  3)  that  arcs  of  great  cir- 
cles between  the  pole  and  circumference  are  likewise  equal. 

Scho.  5.  The  pyramid  or  cone  cut  off  from  another  pyramid 
or  cone  by  a  plane  parallel  to  the  base  is  similar  (VI.  defs.  8 
and  27)  to  the  whole  pyi-amid  or  cone. 

Prop.  XIX. — Theor. — If  the  altitixde  of  a  parallelopipedy 
and  the  length  and  p  rpendicular  breadth  of  its  base  be  all 


170  THE   ELKMKNTS   OF  [bOOK   VI. 

d'vided  into  parts  equal  to  one  another^  the  continued  p  oduct 
of  the  numler  of  parts  in  the  three  lines  is  the  nvmhtr  ofctiba 
contained  in  the  parcdldopiped,  each  cube  havinrj  the  side  of  its 
base  equal  to  one  (f  the  parts. 

P'irst,  suppose  the  ])nralk'loi)ipefl  to  1)8  rcctanccular,  Tlien 
])laiies  panvlU'l  to  the  base  i)assiiii;  tliroiiu'h  the  points  ot'seetion 
of  the  nh.itude  will  evidently  divide  the  so'id  into  as  many- 
equal  solids  as  there  ave  parts  in  the  altitude;  and  each  of 
tliese  partial  solids  will  l)e  composed  of  as  many  cuius  as  the 
base  contains  squares,  each  equal  to  a  base  of  one  of  the  cubes. 
13ut  (I.  23,  cor.  4)  the  number  of  these  squares  is  the  product 
cf  the  length  and  breadth  of  the  base;  and  hence  the  entire 
number  of  cubes  will  be  equal  to  the  product  of  the  three 
dimensions,  the  length,  breadth,  and  altitvide. 

If  the  base  be  not  rectanj,ndar,  its  area  (I.  23,  cor.  5)  will  be 
the  product  of  its  lenjxth  and  j.erpcndicular  breadth;  and  it  is 
evident  that  the  product  of  this  by  the  altitude  will  be  the 
number  of  cubes  as  before. 

Lastly,  if  the  iusistiuii  lines  be  not  perpendicular  to  tlie  base, 
still  the  oblique  parallelopi])ed  is  equal  (VI.  15,  and  cor.  1)  to 
a  rectano-ularone  of  the  same  altitude;  and  therefore  the  num- 
ber  of  cubes  will  be  found  as  before,  by  multiplying  the  area 
of  the  base  by  the  altitude. 

Cor.  1.  Hence  it  is  evident  that  the  volume,  or  numerical 
solid  C07itent,  or,  as  it  is  also  called,  the  solidify,  of  a  parallelo- 
piped  is  the  product  of  its  altittule  and  the  area  of  its  base 
both  expressed  in  numbers;  and  it  is  ])lain  that  the  same 
holds  in  i-egard  to  any  prism  whatever,  and  also  in  regard  to 
cylinders. 

Cor.  2.  The  content  of  a  pyramid  or  cone  is  found  by  multi- 
plying the  area  of  the  base  by  the  altitude,  and  tak'ug  a  third 
of  the  product.  For  (VI.  17)  a  pyramid  is  a  third  part  of  a 
prism,  and  (VI.  17)  a  cone  a  third  jtart  of  a  cylinder,  of  the 
same  base  and  altitude. 

An  easy  method  of  comjmting  the  content  of  a  truncated 
pyramid  or  cone,  that  is,  the  frustum  which  remains  when  a 
part  is  cut  from  the  top  by  a  plane  parallel  to  the  base,  may  be 
thus  investigated  by  the  help  of  algebra.  The  solid  cut  off  is 
(VI.  18,  scho.  3)  similar  to  the  whole;  and  tlierefore  the  areas 


BOOK  VI.]         EUCLID  AND  LEOENDRE.  171 

of  their  bases  will  be  proportional  to  the  squares  of  their  cor- 
respondino^  dimensions,  and  consequently  to  the  squares  of 
tlieir  altitudes.  Hence  puttins^  V  to  denote  the  volume  or  con- 
tent of  the  frustum,  11  and  B  the  altitudes  and  base  of  the 
whole  solid,  and  h  and  h  those  of  the  solid  cut  oif,  if  we  put 
^IP  to  denote  B,  since  B  :  6  :  :  H*  :  A",  or  B  :  6  : :  qW  \  qh^^ 
we  shall  have  (IV.  2,  cor.  1)  b  =  qh'^-^  and  therefore (VI.  19, cor. 
2)  the  contents  of  the  whole  cone  and  the  part  cut  off  are  equal 
respectively  to  ^^■IP  and  ^qlf  \  wherelbre  \  =  ^q(\¥—h')^  ov^ 
by  resolvinnf  the  second  member  into  factors,  V=:i'/(H-+ HA 
+  /r-)  {\l-h)=^{qW+q\lh  +  qh')  (H-A).  Now  ^H^  is  equal 
to  B,  qli'  to  6,  qWh  to  a  mean  proportional  between  them, 
and  II — h  to  the  height  of  the  fi'ustum.  Hence,  to findthe  con- 
tent of  a  truncated  pyramid  or  cone,  add  to<jet/ter  the  areas  of 
its  two  bases  aud  a  mean  proportional  between  theni^  mxdtiply 
the  sum  by  the  height  of  the  frustum,  anddivide  the  productby  3, 
This  admits  of  convenient  modifications  in  particular  cases. 
Thus,  if  the  bases  be  squares  of  which  S  and  s  are  sides,  and  if 
a  be,  the  altitude  of  the  frustum,  we  shall  have 

V=ia(S'+Ss-f-0=^»(3Ss+S-— 2Ss+5'); 
or,  Vz^ia{3S5+(S— s)^}=a{S5+i(S— s)'}. 

Hence,  to  find  the  content  (f  the  frustum  of  a  square  pyramid^ 
to  the  rectangle  under  the  sides  of  its  bases  add  a  third  of  the 
square  of  their  difference,  and  mxdtiply  the  sum.  by  the  height. 
It  would  be  shown  in  like  manner  (V.  25,  scho.  and  VI.  IV, 
cor.  2),  that  if  II  and  r  be  the  radii  of  the  bases  of  the  frustum 
of  a  cone,  and  a  its  altitude, 

N  '^Za{Vxr\-\^—rY\. 

Solidity  of  cylinder,  3  x  11"  x  a. 

Solidity  of  cone,  R^Xflf. 

Solidity  of  sphere,  is  4R'. 

Solidity  of  spherical  sector,  is  211' X  or. 

Solidity  of  spherical  segment,  when  it  has  two  bases,  is 

f  R'4-r)xa+ia'; 

and  when  it  has  but  one  base, 

^R-'x«  +  ^a'. 

Cor.  3.  The  content  of  a  polyhedron  may  be  found  by  divid- 


172 


THE   ELEMENTS   OF 


[book  VI. 


ing  it  into  pyramids,  and  adding  togetlicr  tlieir  contents.  The 
division  into  pyramids  may  be  matle  either  by  phuies  passini^ 
throng!)  tlie  vertex  of  one  of  the  soliil  angles,  or  by  planes 
passing  throngh  a  point  within  tlie  body. 

Piiop.  XX. — Theor. —  The  surfaces  of  two  similar  polyhe- 
drons may  be  divLdedinto  the  same  number  cf  similar  triangles 
similarly  situated. 

This  fullows  immediately  from  the  definition  (VT.  def.  8)  of 
similar  bodies  bounded  by  planes,  if  the  sides  or  faces  of  the 
polyhedron  be  triangles;  and  any  face  in  the  one,  and  the  cor- 
responding face  in  the  other,  which  are  not  triangles,  are  yet 
similar,  and  may  be  divided  (I.  20)  into  the  same  number  of 
similai-  tiiangles  similarlj'  situated. 

Cor.  Hence  it  would  be  shown,  as  in  tlie  fourteenth  proposi- 
tion of  the  fifth  bo()k,  that  the  surfaces  of  the  polyhedions  are 
proportional  to  any  two  of  their  ^similar  triangles ;  and  there- 
fore they  are  to  one  another  in  the  duplicate  ratio  of  the 
homologous  sides  of  those  triangles,  that  is,  of  the  edges  or  in- 
tersections of  the  similar  planes.  Hence  also  the  surfaces  are 
proportional  (V.  14,  coi'.  2)  to  the  squares  of  the  edges. 

Prop.  XXI. — Theor. —  Triangular  pyramids  are  similar,  if 
two  faces  in  one  of  them  be  similar  to  twj  faces  in  the  other^ 
each  to  each,  and  their  inclinations  equ<d. 

Let  ABC,  abc  be  the  bases,  and  D,  d  the  vertices  of  two  tn- 
angular  pyramids,  in  which  ABC,  DBC  are  respectively  similar 
to  abc,  dbc,  and  the  inclination  of  ABC,  DBC  equal  to  that  of 
abc,  dbc ^  the  pyi'amids  are  similar. 

To  demonstrate  this,  it  is  sufficient  to  show  that  the  triangles 
j^  AI>D,  ACD  are  similar  to  ahd^ 

acd,  i\)Y  then  the  solid  angles 
(Vi.  9,  cor.  3)  will  be  equal,  each 
to  each,  and  (VI.  def  8)  the  pyr- 
amids similar.  Since  the  plane 
angles  at  B  and  b  are  equal,  the 
inclinations  of  ABC,  DBC,  and 
of  vhc,  dbc,  are  (VI.  9,  cor,  2)  equal ;  therefore  ABl),  abd  are 
equal.     Then  (hyj..)  DB  :  BC  :  ;  c/i  :  ic,  and  BC  ;  BA  :  :  dc  : 


BOOK   VI.]  EUCLID    AND    LEGENDRE.  173 

ba/  whence,  ex  ceqnn,  DB  :  BA  :  :  clb  :  ha  ;  and  therefore  (V. 
6)  the  trianirles  ABD,  al>d  are  equiangular,  and  consequently 
Bimilar  ;  and  it  would  be  proved  in  the  same  maimer  that 
ACD,  acd  are  similar.  Therelure  (VI.  def.  8)  the  pyramids 
are  similar. 

Cor.  Hence  triangular  pyramids  are  similar,  if  three  faces  of 
one  of  them  be  respectively  similar  to  three  faces  of  the  other. 

In  the  triangular  pyramids  ABCD,  ahcd  (see  the  preceding 
figure),  let  the  faces  ABC,  ABD,  DBC  be  similar  to  ahc^  abd, 
doc,  each  to  each ;  the  pyramids  are  similar. 

For  (V.  def  1)  AD  :  DB  ■.-.  ad  :  dh,  and  DB  :  DC  : :  cf5  : 
dc  ;  whence,  ex  aequo,  AD  :  DC  : :  ad  '.  dc.  Also  DC  :  CB 
: :  dc  :  cb,  and  CB  :  CA  :  :  cb  :  ca ;  whence,  ex  ceqi/o,  DC  : 
CA  :  :  dc  :  ca  ;  and  therefore  (V.  5)  the  triangles  ADC,  adc 
are  equiangular,  and  (VI.  9,  cor.  3,  and  def.  8)  the  pyramids 
are  similar. 

Prop.  XXII. — Tiieor. — Similar  polyhedrons  may  be  divid- 
ed nfo  the  same  riumfer  of  triangular  pyratnids,  similar,  each 
to  each,  and  similarly  situated. 

Let  ABCDEFG  and  ahcdefg  be  similar  polyhedrons,  having 
the  solid  angles  equal  which  are  marked  with  the  corresponding 
large  and  small  letters ;  they  may  be  divided  into  the  same 
number  of  similar  triangular  jjyramids  similarly  situated. 

The  surfaces  of  the  polygons  may  be  divided  (VI.  20)  into 
the  same  number  of  sim  lar  triangles,  similarly  situated;  then 
planes  passing  through  any  two  corresponding  solid  angles,  A, 
a,  and  through  the  sides  of  all  these  triangles,  except  those 


forming  the  solid  angles.  A,  a,  will  divide  the  polyhedrons  into 
triangular  pyramids,  similar  to  one  another,  and  similarly  sit- 
uated. 


174 


TnE   ELEMENTS   OF 


[book  VT. 


The  pyramids  thus  formocl  have  each  one  solid  anole  at  the 
common  vertex  A  or  a  ;  and  these  solid  angles  may  be  of  three 
classes:  Is^  those  which   have  two  of  their  faces  coincidins 
with  faces  of  one  of  the  polyhedi'ons;  2f/,  those  which   have 
only  one  face  coinciding ;  and  3o?,  those  which  lie  wholly  with- 
in the  solid  angle  A  or  a.     Now  those  of  the  first  kind  in  one 
of  the  polyhedrons  are  similar  to  the  corresponding  ones  in  the 
other,  by  the  corollary  to   the  twenty-first  proposition  of  this 
Look  ;    and    those    of    the   second    kind    by    the   twenty-first 
From  the  polyhedrons  take  two  of  these  similar  ])yramids,  and 
the  remaining  bodies  will  be  similar,  as  the  boundaries  common 
to  them  and  the  pyramids  are  (VI.  21,  and  cor.)  similar  trian- 
gles ;  and  their  other  boun<laries  are  similar,  being  faces  of  the 
proposed  polyhedrons.     Also  the  solid  angles  of  the  remaining 
bodies  are  equal,  as  some  of  them  are  angles  of  the  primitive 
polyhedrons,  and  the  rest  are  either  trihedral  angles  which  are 
contained  by  equal  plane  angles,  or  may  be  divided  into  such. 
From  these  remaining  bodies  other  similar  triangular  pyramids 
may  be  taken  in  a  similar  manner,  and  the  process  may  be  con- 
tinued till  only  two  similar  triangular  pyramids  remain  ;  and 
thus  the  polyhedrons  are  resolved  into  the  same  number  of 
eimilar  triangular  pyramids. 

Prop.  XXIII. — Pitoc. —  To  find  the  diameter  of  a  given 
sphere. 

Let  A  he  any  point  in  the  surface  of  the  civen  sphere,  and 
take  any  three  jjoints  B,  C,  I)  at  equal  distances  from  A.  De- 
ecribe  the  triangle  hcd  having  he  equal  to  the  distance  or  chord 
BC,  cd  equal  to  CD,  and  hd  to  BD.     Find  e  the  center  of  the 


circle  described  about  hcd^  and  join  he;  draw  a^perpendicular 
to  he^  and  make  ha  eqtial  to  BA  ;  draw  ^perpendicular  to  6a, 
and  af  is  equal  to  the  diameter  of  the  sphere. 


BOOK  VI."]         EUCLID  AND  LEGENDRE.  175 

Conceive  a  circle  to  be  described  through  BCD,  and  E  to  be 
its  center;  that  circle  will  evidently  be  the  section  of  the 
sphere  by  a  plane  throui;h  B,  C,  D ;  and  it  will  be  equal  to 
the  circle  described  about  bed.  Conceive  the  diameter  AEF 
to  be  drawn,  and  liA,  BE,  BF  to  be  joined.  Then,  in  the 
right-angled  triangles  ABE,  abe^  the  sides  AB,  BE  are  respect- 
ively equal  to  ab^  be,  and  therefore  (I.  24,  cor.  2)  the  angles  A, 
a  are  equal. 

Again :  in  the  right-angled  triangles  ABF,  abf,  the  angles 
A,  a  are  equal,  and  also  the  sides  AB,  ab/  hence  (I.  14)  the 
Bides  AF,  a/' are  equal;  that  is,  a/ is  equal  to  the  diameter  of 
the  sphere. 

Pkop.  XXIV. — TiiEOR. —  T/ie  angle  of  a  spherical  triangle  is 
the  angle  formed  by  the  tangents  of  the  arcs  forming  the 
fpherical  angle,  and  is  measured  by  the  arc  of  a  great  circle 
described  from  the  vertex  as  apcle,  and  intercepted  by  the  sides, 
produced  if  necessary. 

Let  BAC  be  a  spherical  angle  formed  by  the  arcs  AB  and 
AC,  then  it  is  the  same  as  the  angle  EAD  fonncd  by  the  tan- 
gents EA  ajid  DA,  and  is  measured  by  the  arc 
of  a  great  circle  intercepted  by  the  arcs  AB 
and  AC,  produced  if  necessary.  The  tangents 
AE  and  AD  are  both  jierpendicular  to  the 
common  diameter  AH  (HI.  12),  and  being  in 
the  same  planes  as  the  arcs  AB  and  AC,  form 
an  angle  EAD  equal  to  the  spherical  angle 
BAC.  ^ 

Again :  the  radii  FB  and  FC  of  the  great  circle  described 
from  the  vertex  as  a  pole,  being  in  the  same  planes  as  the  arcs 
AB  and  AC,  and  perpendicular  to  AH,  are  parallel  with  AE 
and  AD  respectively,  hence  the  angle  EAD  is  equal  to  the 
angle  BFC.  But  the  angle  BFC  is  measured  by  the  arc  BC 
(I.  def.  19) ;  therefore,  also  (I.  ax.  1),  the  spherical  angle  BAC 
is  measured  by  the  aic  BC. 

^ho.  The  angles  of  spli(  rical  triangles  may  be  compared  to- 
gether by  means  of  the  arcs  of  great  circles  described  from 
their  vertices  as  poles  aid  included  between  the  arcs  forming 


176  THE   ELEMENTS   OP  [bOOK   TI. 

the  angles,  and  it  is  easy  to  make  a  spherical  angle  equal  to  a 
given  angle. 

Cor.  1.  If  from  the  vertices  of  the  three  angles  of  a  spherical 
triangle  as  poles,  arcs  be  described  forming  a  spherical  tiiangle, 
then  the  veitices  of  the  angles  of  this  second  triangle  will  be 
respectively  poles  of  the  sides  of  the  first,  and  each  angle  will 
be  measured  reciprocally  by  a  semicircumference  less  the  side 
of  the  other  triangle  opposite  to  the  angle. 

Because  A,  B,  and  C  are  poles  respectively  of  the  arcs  FE, 
ED,  and  DF,  the  distances  of  the  poles  from  the  extremities  of 
their  respective  arcs  are,  in  each  case,  a  quadrant ;  hence  the 


extremities  of  the  ai'cs  FE,  ED,  and  DF  are  respectively  re- 
moved the  length  of  a  quadrant  from  the  extremities  of  the 
arcs  AB,  BC,  and  AC ;  therefore  the  extremities  of  the  former 
arcs  are  the  poles  of  the  latter  arcs,  each  to  each. 

Since  A  is  the  pole  of  the  arc  GH,  the  angle  BAG  is  meas- 
ured by  that  arc  (VI.  24),  and  F  being  the  pole  of  AH,  FH  is 
a  quadrant,  and  E  being  the  pole  of  AG,  GE  is  a  quadrant; 
hence  Fllf  GEO  semicircumference;  but  FH  +  GEoFE-f 
GIl,  or  the  arc  GH,  which  measures  the  angle  BAG,  is  equiva- 
lent to  a  semicircumference  less  the  arc  FE.  In  like  manner, 
the  angle  ABC  can  be  shown  to  be  measured  by  a  semicircum- 
ference less  the  arc  DE,  and  the  angle  ACB  to  be  measured  by 
a  semicircumference  less  the  arc  DF.  And,  reciprocally,  the 
angle  FDE  is  measured  by  the  arc  LO.  But  LO  +  BCoLC 
4-B0=O  semiciivumfcrence;  hence  L0=O  semicircumference 
minus  BC  ;  and  a  similar  condition  can  be  shown  for  the  other 
ambles  of  the  triauijle  FED. 


BOOK   VI,]  EUCLID   AND   LEGENDRE.  177 

Cor.  2.  As  each  angle  of  a  spherical  triangle  is  less  than  two 
right  angles,  the  three  angles  are  less  than  six  right  angles. 
And  as  the  sum  of  the  sides  of  a  spherical  triangle  is  less  than 
the  cij'cumference  of  a  great  circle,  and  the  angles  being  meas- 
ured (VI.  24,  cor.  1)  by  three  semicircumferences  less  the  three 
sides  of  the  polar  triangle,  taking  away  the  sides,  we  have 
the  remainder  greater  than  one  semicircumference — or  the 
three  angles  greater  than  two  right  angles.  Hence  the  angles 
of  a  spherical  triangle  vary  between  two  right  angles  and  six 
right  angles,  without  reaching  either  limit ;  therefore  two  an- 
gles given  can  not  determine  the  third. 

12 


END   OF  BOOK   SIXTH, 


THE  ELEMENTS  OF  PLANE  TRIGONOMETRY. 

DEFIXITIOXS. 

1.  TRiGOJfOMETRY  is  the  practical  application  of  geometrical 
principles  for  the  investigation  of  ratios  of  the  sides  of  triangles 
in  connection  with  the  magnitudes  of  their  angles.  Yor perspi- 
cuity, the  vertex  of  the  angle  is  placed  in  the  center  of  a  circle, 
and  the  arc  of  the  circumference  intercepted  by  the  sides  con- 
taining the  angle  is  used  as  2i  measure  of  the  angle  (L  def  19). 
Let  a  straight  line  be  supposed  to  move  around  a  fixed  point ; 
it  will  make  with  a  stationary  line  angles  which  will  vary  as  the 
line  is  moved,  and  when  it  has  passed  around  until  it  coincides 
with  the  stationary  line  from  which  it  is  supposed  to  have 
started,  it  will  have  gone  over  the  magnitude  of  four  right 
angles  (I.  9,  cor.),  the  extremity  of  the  movable  line  will  trace 
the  circumference  of  a  circle,  and  the  successive  arcs  intercept- 
ed between  the  movable  and  stationary  lines  will  give  the 
magnitude  of  the  angles  (I.  def  19). 

2.  For  the  purposes  of  calculation,  a  right  angle  is  divided 
into  an  arbitrary  number  of  equal  parts;  each  one  of  these 
parts  is  subdivided  into  other  equal  parts,  and  each  part 
of  this  subdivision  undergoes  a  second  subdivision  of  equal 
parts,  and  when  particular  nicety  and  precision  are  desired, 
there  is  a  third  subdivision  of  equal  parts.  Thus,  the  first 
division  of  a  right  angle  is  into  degrees.  Among  the  English 
mathematicians,  a  right  angle  has  ninety  degrees,  which  divi- 
sion is  derived  from  Greek  works,  and  has  great  antiquity, 
being  used  by  the  remotest  ancient  mathematicians  and  astron- 
omers of  whom  we  have  any  account.  Among  some  modem 
French  mathematicians,  a  right  angle  is  divided  into  one  hun- 
dred degrees,  which  centesimal  division  is  continued  through- 
out all  the  various  subdivisions.  But  in  the  Greek  division, 
which  is  more  generally  used  on  account  of  the  great  facility 
with  which  360  can  be  subdivided,  each  degree  has  sixty  equal 


PLANE   TKIGONOMETET,  179 

parts  called  minntes^  each  minute  has  sixty  equal  parts  called 
seconds,  and  each  second  is  sometimes  subdivided  in  deci- 
mal parts,  and  thus  the  most  extreme  minuteness  can  be  ob- 
tained. 

3.  The  symbols  for  abbreviation  in  the  expression  of  the 
value  of  angles  are  as  follows:  °,  \  "\  thus,  60°,  15',  "lb"  are 
read,  sixty  degrees,  fifteen  minutes,  and  twenty-five  seconds. 

4.  The  reason  why  the  right  angle  is  assumed  for  division  is 
because  that  angle  preserves  an  inversion  between  every  angle 
less  than  a  right  angle  and  its  complement  (L  def  20),  and  a 
similarity  between  every  angle  greater  than  a  right  angle  and 
its  supplement  (I.  def  20) ;  thus,  in  the  first  case,  the  functions 
of  the  angle  are  inverted  in  respect  to  its  complement,  and  in 
the  latter  case  the  functions  are  the  same  in  respect  to  its 
supplement,  as  will  more  readily  be  seen  by  the  seventh  defi- 
nition and  following  corollaries. 

5.  The  straight  line  drawn  from  one  extremity  of  an  arc, 
perpendicular  to  the  diameter  passing  through  the  other  ex- 
tremity, is  the  sine  of  the  angle  measured  by  it ;  and  the  part 
of  that  diameter  intercepted  between  the  sine  and  the  arc  is 
the  versed  sine  of  the  angle  which  it  measures. 

6.  If  a  straight  line  touch  a  circle  at  one  extremity  of  an  arc, 
the  part  of  it  intercepted  between  that  extremity  and  the 
diameter  produced,  which  passes  through  the  other,  is  the  tati- 
gent  of  the  angle  which  it  measures;  and  the  straight  line 
drawn  from  the  center  to  the  remote  extremity  of  the  tangent 
is  the  secant  of  the  angle. 

Y.  The  cosine  of  an  angle  is  the  sine  of  its  complement.  In 
like  manner,  the  coversed  sine,  cotangent,  and  cosecant  of  an 
angle  are  respectively  the  versed  sine,  tangent,  and  secant  of 
its  complement. 

The  sine,  versed  sine,  tangent,  and  secant  may  be  denoted 
by  the  abbreviated  expressions,  sin,  versin  (or  vs.),  tan,  and 
sec/  and  the  cosine,  coversed  sine,  cotangent,  and  cosecant, 
by  cos,  coversin  (or  covs),  cotan,  and  cosec.  For  the  sake  of 
eimplicity,  the  radius  of  the  circle  employed  for  comparing  dif- 
ferent angles  is  generally  taken  in  investigations  as  unity; 
when  this  is  not  done,  it  is  denoted  by  its  initial  letter  R. 

The  sides  of  a  triangle  are  often  conveniently  denoted  by  tha 


180 


THE    ELEarENTS    OF 


Bmall  letters  corresponding  to  the  capital  ones  placed  at  the 
opposite  angles.  Thus,  a  denotes  the  side  opposite  to  the  an- 
gle A,  etc.  To  prevent  ambiguity,  we  may  read  A,  B,  C ; 
angle  A,  angle  B,  angle  C ;  while  a,  6,  c  may  be  called  side  a, 
side  b,  side  c. 

To  illustrate  the  foregoing  definitions,  let  C  be  the  center  of 
a  circle,  and  AB,  DE  two  diameters  perpendicular  to  each 
other.  Through  any  point  F  in  the  circumference  draw  the 
diameter  FL;  draw  FG  perpendicular  to  AB,  and  FI  to  DE; 
through  A  draw  AH  perpendicular  to  AB,  and  therefore  (III. 
8,  cor.)  touching  the  circle  in  A  ;  and  let  it  meet  LF  produced 
in  H ;  and,  lastly,  draw  DK  perpendicular  to  DE,  meeting  FL 


H 

D 

/^ 

~^^ 

~y 

!^ 

T^N. 

( 

/ 

r 

^ 

\/ 

Q 

C 

\  ; 

k 

J 

^ 

N 

\. 

V 

E 


produced  in  K.  Then  the  arc  AF  contains  the  same  number 
of  degrees,  etc.,  as  the  angle  ACF ;  and  FG  is  the  sine  of  this 
angle ;  FI,  or  its  equal  CG,  the  cosine ;  AG  the  versed  sine, 
and  DI  the  coversed  sine ;  AH  the  tangent,  and  CH  the  secant  j 
DK  the  cotangent,  and  CK  the  cosecant. 

From  these  definitions  we  derive  the  following  corollaries : 

Cor.  1.  The  sine,  of  an  angle  ACF  is  half  the  chord  of  double 
the  arc  measuring  it.  For  if  FG  be  produced  to  meet  the  cir- 
cumference in  N,  FN  is  bisected  (III.  2)  in  G,  and  (III.  17)  the 
arc  FAN  in  A. 

Cor.  2.  The  sine  of  the  right  angle  ACD  is  the  radius  CD. 

Cor.  3.  If  AF  bo  half  of  AD,  and  consequently  ACF  half  a 
right  angle,  the  tangent  AH  is  equal  to  the  radius.  For  A  be- 
ing a  right  angle,  II  must  be  half  a  right  angle,  and  (I.  1,  cor. 
2)  AH  equal  to  AC. 

Cor.  4.  Put  the  angle  =  A,  and  the  radius=:l.  Then  (I.  24, 
cor.    1)    FG-+CG'=CF^;  that   is,   siu=A+cos'A  =  l.     In   like 


PLANE    TRIGONOMETRY. 


181 


manner,  we  find  from  the  right-angled  triangles  CAH,  CDK, 
that  Cir=CA^+AH^  and  CK^  =  CD'+DK^;  that  is,  sec^A= 
l-ftan^A,  and  cosec^'A^l+cot'^A. 


Cor.  5.  In  the  similar  triangles  CGF,  CAH,  CG  :  CF,  or 
CA  :  :  CA  :  CH  ;  that  is,  the  cosine  of  an  angle  is  to  the  ra- 
dius as  the  radius  to  its  secant.  Hence  also  (V.  9,  cor.)  CG.CH 
=:CA^;  that  is,  cosAsecA  =  l.  It  would  be  found  in  like 
manner  from  the  triangles  GIF,  CDK,  that  sin  A  cosecA=l, 
CI  being  equal  to  the  sine  FG. 

Cor.  6.  In  the  same  triangles  CGF,  CAH,  the  cosine  CG  is 
to  the  sine  GF  as  the  radius  CA  to  the  tangent  AH ;  whence 
(V.  8)  cos  A  tanA=sin  A.  The  triangles  CIF,  CDK  give  in 
like  manner  sin  A  cot  A  =  cos  A. 

Cor.  7.  The  radius  is  a  mean  proportional  between  the  tan- 
gent of  an  angle  and  its  cotangent.  For  the  triangles  CAH, 
CDK  are  similar;  and  therefore  HA  :  AC  :  :  CD,  or  CA  : 
DK.     Hence  (V.  9,  cor.)  tan  A  cot  A=:l. 

Cor.  8.  The  sine  of  an  angle,  and  the  sine  of  its  supplement 
are  equal.  So  likewise  are  their  cosines,  tangents,  cotangents, 
secants,  and  cosecants. 

Let  ACF  be  an  angle,  FG,  AH  its  sine  and  tangent,  and 
CG,  DK  its  cosine  and  cotangent.  Make  the  angle  BCM  equal 
to  ACF ;  draw  the  perpendicular  MO  ;  and  produce  MC  both 
ways  to  meet  HA,  KD  produced  in  P  and  Q.  Then  (I.  def. 
20,  and  I.  9)  the  angles  BCM,  ACM,  or  ACF,  ACM  are  sup- 
plements of  each  otlier ;  as  are  also  the  arcs  BM,  AM,  or  AF, 
AM,  since  (HI.  16)  BM,  AF  are  equal.  Now  the  triangles 
CGF,  COM  are  equiangular,  and  have  the  sides  CF,  CM  equal ; 


182 


THE   ELEMENTS   OF 


Q 

T) 

/ 

! 

0 

\ 

/ 

^ 

\ 

V 

C 

\ 

J 

4 

H 


therefore  (I.  14)  MO  is  equal  to  FG,  and  CO  to  CG;  and  MO, 

FG  are  the  sines  of  ACM,  ACF,  and  CO, 
CG  their  cosines.  Again  :  the  triangles 
ACP,  ACH  are  equiangular,  and  have 
AC  common;  therefore  (I.  14)  AP  is 
equal  to  AH,  and  CP  to  CH ;  and  AP, 
AH  are  the  tangents,  and  CP,  CH  the 
secants  of  ACM,  ACF.  In  like  manner  it 
would  be  proved,  by  means  of  tlie  trian- 
gles CDQ,  CDK,  that  DQ,  the  cotangent 

of  ACM,  is  equal  to  DK,  the  cotangent  of  ACF,  and  that  their 

cosecants  CQ,  CK  are  equal. 


PROPOSITIOXS. 

» 

Pkop.  I. — TnEOR. — T/i  a  right-ancjled  triangle  the  hypothe- 
7iuse  is  to  either  of  the  legs  as  the  radius  to  the  sine  of  the  an^ 
gle  opposite  to  that  leg,  or  to  the  cosine  of  the  adjacent  angle  ; 
(2)  either  of  the  legs  is  to  the  other  as  the  radius  to  the  tangent 
of  the  angle  opposite  to  the  latter  ;  and  (3)  either  of  the  legs  is 
to  the  hypothenuse  as  the  radius  to  the  secant  of  the  contained 
angle. 

Let  ABC  be  a  triangle,  right-angled  at  C  ;  then  (1)  c  :  h  :: 
R  :  sin  B,  or  cos  A ;  (2)  a  :  6  :  :  R  :  tau  B ;  and  (3)  a  :  c  : : 
R  :  sec  B. 

From  B  as  center,  with  any  radius,  describe  an  arc  cutting 
AB,  BC  in  D,  E;  and  through  D,  E  draw  (I.  8  and  V)  DF, 
^     EG   perpendicular  to   BC.     Then    (Trig. 
defs.  5  and  6)  FD,  EG,  and  BG  are  re- 
spectively the  sine,  tangent,  and  secant  of 
the  angle  B ;  and,  since  C  is  a  right  angle, 
A  and  B   (Trig,  def  4)  are  complements 
of  each  other;  and  therefore  (Trig,  def  7) 
sinB=cosA.     Ao-ain  :  since  the  an^le  B 
is  common  to  the  triangles  ABC,  DBF, 
GBE,  and  the  angles  at  C,  F,  E  right  angles,  these  triangles  (I. 
20,  cor.  5)  are  equiangular. 

Hence  (V.  3)  in  the  triangles  ABC,  DBF, 

BA  :  AC  : :  BD  :  DF ;  that  is,  c  :  ^<  : :  R  :  sin  B,  or  cos  A. 
Again :  (V.  3)  in  the  triangles  AP.C,  GBE, 


PLANE   TRIGONOMETRY. 


183 


BC  :  CA  : :  BE  :  EG ;  that  is,  a  :  5  : :  R  :  tan  B ;  and 
BC  :  BA  : :  BE  :  BG ;  that  is,  a  :  c  : :  R  :  sec  B. 
Cor.  Hence  (V.  10,  cor.)  RJ=c  sinB=c  cos  A;  that  is,  the 
'product  of  either  leg  and  the  radius  is  equal  to  the  product  of 
the  hypothenuse  and  the  sine  of  the  angle  opposite  to  that  leg^ 
or  of  the  hypothenuse  and  the  cosine  of  the  adjacent  angle. 
When  R=:l,  this  becomes  simply  b=c  siu  B=c  cos  A.  Again  : 
J=a  tan  B  and  c=:a  sec  B. 


Prop.  II. — Theor. —  The  sides  of  a  plane  triangle  are  pro- 
portional to  the  sines  of  the  opposite  angles. 

Let  ABC  be  any  triangle  ;  then  a  :  b  :  :  sin  A  :  sin  B  ;  a  : 
c  :  :  sin  A  :  sin  C ;  and  b  :  c  :  :  sin  B  :  sin  C. 

Draw  AD  perpendicular  to  BC;  then  AD  is  a  leg  of  each  of 

the    right-angled    triangles     ADB, 
ADC;  and  therefore  (Trig,  1,  cor.) 
R.AD=AB  sin  B,  and  R.AD^AG 
sin  C.     Hence  (I.  ax.  1)  AB  sin  B  = 
AC    sin  C,    or    c   sin  B  =  5    sin  C; 
whence  (V.  10,  cor.)  b  :  c  :  :  sin  B 
:  sin  C ;  and,  by  drawing  perpendic- 
ulars from  B  and  C  to  the  opposite  sides,  it  would  be  proved 
in  a  similar  manner  that  a  :  c  :  :  sin  A  :  sin  C,  and  a  :  b  ::  sin 
A  :  fin  B. 

Cor.  From  B  as  center  M'ith  BA  as  radius,  describe  an  arc 
AD  ;  and  from  C  as  center,  with  an  equal  radius,  describe  an  arc 
EF.  Draw  AG,  EH  perpendicular  to  BC  ;  these  (Trig.  def.  5) 
are  respectively  the  sines  of  B  and  C  to  equal  radii.  Then 
the  triangles  AGC,  EHC  are  equiangular,  the  angle  at  C  being 
common,  and  the  angles  at  G  and  H  right  angles.  Hence  (V. 
3)  CA  :  AG  :  :  CE,  or  (const.)  AB  :  EH ;  and,  alternately, 
CA  :  AB  :  :  AG  :  EH ;  that  is,  5  :  c  :  :  sin  B  :  sin  C. 

The  demonstration  is  simplified  by  taking,  as  here,  one  of  the 
sides,  AB  or  AC,  as  radius.  This,  however,  is  not  essential,  as 
arcs  may  be  described  from  B  and  C  as  centers,  with  equal 
radii  of  any  magnitude,  and  their  sines,  and  a  perpendi-cular 
from  A  to  BC  being  drawn,  the  proof  will  be  readily  obtained. 
Scho.  From  one  of  the  foregoing  analogies  we  have,  by  in- 
version, c  :  J  :  :  sin  C  :  sin  B.      If  C  be  a  right   angle,  this 


184  THE   ELEMENTS    OF 

(Trig.  def.  cor.  2)  becomes  c  :  Z>  :  :  R  :  sin  B,  as  in  Prop.  L 
The  first  part,  therefore,  of  that  proposition  is  a  particular  case 
of  this  one. 

Prop.  ITT. — Theor. — The  sum  of  any  Uco  sides  of  a  trian- 
gle is  to  their  difference  as  the  tangent  of  half  the  snm  of  the 
angle  opposite  to  those  sides  is  to  the  tangent  of  half  their  dif- 
ference. 

].et  ABC  be  a  triangle,  a,  h  any  two  of  its  sides,  of  which  a 
is  the  greater,  and  A,  B  the  angles  opposite  to  them ;  then 
a+b  :  a-h  ::  tan  i(A  +  B)  :  tan  i(A— B). 

From  C  as  center,  with  the  greater  side  a  as  radius,  describe 

the  circle  DBE,  cutting  AC  produced 
in  D  and  E,  and  BA  produced  in  F ; 
join  BD,  BE,  CF ;  and  draw  EG  par- 
allel to  AB,  meeting  DB  produced  in 
G. 

Then  because  DC  and  CE  are  each 
equal  to  a,  DA  is  equal  to  a+5,  and 
AE  to  a—h. 

Also  (I.  20)  the  exterior  angle  DCB  is  equal  to  A  +  B  ;  and 
DEB,  which  is  at  the  circumference,  is  (III.  10)  half  of  DCB, 
•which  is  at  the  center ;  therefore  DlEB=i(A  +  B). 

Again  (I.  1,  cor.  l) :  the  angle  F  is  equal  to  B  ;  and  (I.  20) 
in  the  triangle  ACF,  the  exterior  angle  A  =  ACF+F=ACF+ 
B;  and  consequently,  ACF= A— B ;  and  (III.  10)  ABE,  or  (I. 
16)  its  equal,  BEG  =  |(A— B). 

Now,  since  (III.  11)  EBD,  being  in  a  semicircle,  is  a  right 
angle,  as  also  (I.  9)  EBG ;  if  a  circle  were  described  from  E  as 
center,  with  EB  as  radius,  DBG  (III.  8,  cor.)  would  touch  it, 
and  (Trig,  def  6)  DB  would  be  the  tangent  of  DEB,  and  BG 
of  BEG;  and  therefore  DB,  BG  will  evidently  be  proportional 
to  the  tangents  of  those  angles  to  any  other  radius. 

Or  strictly,  EB  :  BD  :  :  1  :  tan  DEB  (Trig.  1)  and  (inver.) 

BD  :  EB  ::  tan  DEB  :  1.      Also  (Trig.   1)  EB  :  BG  ::   1  : 

tan  BEG.     Hence,  ex  mquo,  BD  :  BG  :  :  tan  DEB  :  tan  BEG. 

Lastly,  since  BA  (const.)  is  parallel  to  GE,  we  have  (V.  2) 

DA  :  AE  : :  DB  :  BG ;  that  is, 

a-\-b  :  a—b  :  :  tan  |  (A+B)  :  tan  i(A— B). 


PLANE    TRIGONOMETKT. 


185 


Prop.  IV — Tiieor. — hi  a  plane  triangle^  the  cosine  of  half 
the  difference  of  any  two  angles  is  to  the  cosine  of  half  their 
sum,  as  the  sum.  of  the  opposite  sides  to  the  third  side  ;  and  (2) 
the  sine  of  half  the  difference  of  any  two  angles  is  to  the  sine 
of  half  their  sum,  as  the  difference  of  the  opposite  sides  to  the 
third  side. 

Let  ABC  (see  the  last  proposition)  be  any  plane  triangle ; 

then  cosi(A — B)  :  cos^(A  +  B)  : :  a+b  :  c; 
and  sin  \{A — B)  :  sin  -KA+B)  : :  a — b  :  c. 

For  it  was  shown  in  the  preceding  proposition,  that  BED= 
i(A+B),  and  ABE=i(A-B) ;  and  since  DBE  is  a  right  an- 
gle, DBA  is  the  complement  of  ABE,  and  D  of  BED.  But 
(Trig.  2)  in  the  triangle  ABD,  sin  ABD  :  sin  D  : :  AD  :  AB  ; 
that  is,  (Trig.  def.  7)  cosi(A— B)  :  cosi(A+B)  ::  a+b  :  c. 
Again  (Trig,  2)  :  in  the  triangle  ABE,  sin  ABE  :  sin  AEB  : : 
AE  :  AB  J  that  is,  sin  ^(A — B)  :  sin  i(A+B)  : :  a — b  :  c. 

Prop.  V. — Theor. — In  any  plane  triangle  the  sum  of  the 
segments  of  the  base  made  by  a  perpendicular  from  the  vertex, 
is  to  the  sum  of  the  other  sides  as  the  difference  of  those  sides 
to  the  difference  of  the  segments. 

Let  ABC  be  a  triangle,  -and  AD  a  perpendicular  from  the 
vertex  to  the  base ;  the  sum  of  the  segments  BD,  DC  is  to  the 
sum  of  the  sides  AB,  AC,  as  the  difference  of  AB,  AC  to  the 
difference  of  BD,  DC. 


For  (IL  5,  cor.  4)  the  rectangle  under  the  sum  and  difference 
of  AB,  AC  is  equivalent  to  the  rectangle  under  the  sum  and 
difference  of  BD,  DC;  and  therefore  (V.  10,  cor.)  the  sum  of 
BD,  DC  is  to  the  sum  of  AB,  AC,  as  the  difference  of  AB,  AC 
to  the  difference  of  BD,  DC. 

Scho.  If  the  perpendicular  fall  within  the  triangle,  the  seg- 


186  THE    ELEMENTS    OF 

ments  make  up  the  base,  and  their  difference  is  less  than  the 
base;  but  if  the  perpendicular  fall  Avithout  the  triangle,  as  it 
does  (second  fig.)  when  one  of  the  angles  at  the  base  is  obtuse, 
the  base  is  the  difference  of  the  segments,  and  their  sum  is 
greater  than  the  base. 

Peop.  VI. — Theor. —  The  rectangle  under  two  sides  of  a  tri- 
angle is  to  the  rectangle  under  the  excesses  of  half  the  perimeter 
above  those  sides,  as  the  square  of  the  radius  to  the  square  of 
the  sine  of  half  the  contained  angle. 

Let  ABC  be  a  triangle,  and  let  s=i\{a-{-b+c)  ;  then  be  :  (s 
— b)  {s — c)  : :  R^  :  sin'^A. 

Produce  the  less  side  AC  through  C,  making  AD  equal  to 
A  AB ;  join  BD ;  and  draw  AE,  CF  per- 

pendicular, and  CG  parallel  to  BD ; 
then  (I.  24,  cor.  2)  AE  bisects  BD  and 
the  angle  A.  Now  (II.  5,  cor.  4)  the 
rectangle  under  the  sum  and  difference 
of  BC,  CD  is  equivalent  to  tlie  rectangle 
under  the  sum  and  difference  of  BF, 
FD,  that  is,  under  BD  and  twice  EF ;  therefore  the  rectangle 
under  half  the  sum  and  half  the  difference  of  BC,  CD  is  equiva- 
lent to  the  rectangle  BE.EF.     But  (Tkig.  1) 

AB  :  BE  : :  R  :  sin  1  A,  and 

AC  :  CG  or  EF  :  :  R  :  sin i  A;  whence  (IV.  15) 

AC.AB  :  BE.EF,  or  ^  (BC  +  CD).  i  (BC-CD)  : :  R^ :  sin^  A ; 

or,  be  :  |  (a+c — b).  |  (a+5— c)  : :  R^ :  sin  |  A, 
because  CD=c—b  :  and  let  2s=a-\-b-\-c ;  then,  s — a=^  {b-\-c 
— a);s — b—i  {a-\-c — b)  ;  and  s — c—i  {a+b — c)  ;  then  we  have 
be  :  (s—b)  (s—c)  : :  R^  :  suri  A. 

Cor.  Hence,  taking  R  =  l  dividing  the  product  of  the  means 
by  the  first  extreme,  and  exti acting  the  square  root,  we  find 

sin^A  =  Y- ^ 5  and  it  is  plain  that  we  should  find  in  a 

similar  manner,   sin  f  o=\/ —- •,    and    sin  4-  C  = 

y  ac 

V  ab 


PLANE   TRIGONOMETRY.  187 

Prop.  YII. — Theor. —  The  rectangle  under  two  sides  of  a 
triangle  is  to  the  rectangle  under  half  the  perimeter  and  its  ex- 
cess above  the  third  side,  as  the  square  of  the  radius  to  the 
square  of  the  cosine  of  half  the  angle  contained  by  the  two 
sides. 

Let  ABC  be  a  triangle,  and  let  s=^  {a-^-b+c) ;  then  be  :  s 
{s—a)  ::RM  cos^A. 

Produce  tlie  less  side  CA,  through  A,  making  AD  equal  to 
AB  ;  join  BD  ;  draw  AE,  CF  perpen-  d^ 
dicular,  and  AG  parallel  to  BD.  Then 
BD  (I.  24,  cor.  2)  is  bisected  in  E ;  and 
the  angle  BAG  being  (I.  20)  equal  to  the 
two  equal  angles  D  and  ABD,  each  of 
them  is  equal  to  half  the  angle  BAG, 

that  is,  half  the  angle  A  in  the  triangle  ABC;  and  (I.  16) 
GAG  is  equal  to  D.  Now,  it  would  be  shown,  as  in  the  pre- 
ceding proposition,  that  the  rectangle  under  half  the  sum  and 
half  the  diiference  of  DC,  CB  is  equivalent  to  the  rectangle 
BE.EF.  But  (Trig.  1)  AB  :  BE  :  :  R  :  cos  ABE,  or  cos  ^  A  ; 
and  AG  :  AG,  or  EF  : :  R  :  cos  GAG,  or  cos  ^  A ;  whence  (IV. 
15), 
AG.AB  :  BE.EF,  or  i  (DC+GB).  i  (DG-GB) : :  R' :  cos^A ; 

ovbc:  \  {a+b-\-c).  -J  {b-\-c — a)  : :  R^ :  cos^^  A, 
because  DG=6+c.     If  2s  be  the  same  as  last  proposition  : 
be  :  5  (5 — a)  : :  R*^ :  cos''  ^  A. 

Cor.  1.  Hence  we  find,  as  in  the  corollary  to  the  preceding 
proposition,  that 

cos-|A=^  — ^ -\  and  it  would  be  pi'oved  in  a  similar  man- 
ner, that 

costB  =  i/— ^ -i  andcosfG=|/ — —~-. 

Cor.  2.  From  the  sixth  corollary  to  the  definitions  of  trig- 
onometry, it  is  plain,  when  the  radius  is  unity,  if  the  sine  of  an 
angle  be  divided  by  its  cosine,  the  quotient  is  its  tangent. 
Hence,  by  dividing  the  expression  for  the  sine  of  -j  A  in  the 
corollary  to  the  preceding  proposition,  by  the  value  of  its  co- 

,     ,               ,,                  ,      .            ,    .         /is — b)  (s — e) 
Bine  in  the  last  corollary,  we  obtam  tan  ^  A=i/ j^ t —  ; 


188 


THE   ELEMENTS    OF 


and  we  should  obviously  find  in  a  similar  manner,  that 


tan|B: 


{s—a)  {s—c) 


.  As—a)  (^ 
'V      s  {s- 


)  and  tan 


iC=/ 


(s-a)  (s-b) 


-b)      '"'-^""2        y         s(s—c)     ' 

Cor.  3.  By  dividing  the  values  of  tan  ^^B,  tan  4^C,  in  the  pre- 
ceding corollary,  each  by  that  of  tan  |^A,  we  obtain 
tan  iB     5 — a        _  tan  AC     s — a 

TT= V  and   : ,— r= . 

tan  +A     s — o 


tan  ^A     5 — c 


Prop.  VIII. — Prob. —  Given  the  radius  of  a  circle^  and  the 
cosine  of  an  angle^  less  than  a  right  angle ;  to  compute  the 
cosine  of  half  the  angle. 

Let  CD,  the  cosine  of  ACB,  less  than  a  right  angle,  be 
given ;  it  is  required  to  compute  the  cosine  of  its  half 

Draw  the  chord  AB,  and  perpendicular  to  it  draw  CFE ; 
draw  also  FG  parallel  to  BB;  then  (III.  2, 
and  Trig.  defs.  5  and  7)  AF  or  FB  is  the  sine, 
and  CF  the  cosine  of  ACE  tlie  half  of  ACB. 
Also  (V.  2),  DG  is  equal  to  GA,  since  BF  is 
equal  to  FA,  and  DA=2DG;  to  each  of  these 
add  2CD;  then  CA+CDr3  2CG,  and  conse- 
quently CGi=i(CA  +  CD)  ;  or,  if  the  radius  be 
taken  as  unity,  and  the  angle  ACB  be  denoted  by  A,  CG=r| 
(1+cos  A).  Again,  in  the  similar  triangles  ACF,  CFG,  AC  : 
CF  : :  CF  :  CG;  whence  (V.  10,  cor.  2)  CF-r=AC.CG;  that  is, 
cos  2  ^  A=|  (1 +COS  A).  Hence,  to  compute  cos  |  A,  add  1  to 
cos  A,  take  half  the  sum,  and  extract  the  square  root. 

Scho.  This  proposition  and  the  next  afford  means  by  which 
trigonometrical  tables  can  be  computed. 

Prop.  IX.— Theor. — If  A  and  B  be  any  two  angles,  R  : 
cosB::  sinA  :  i  sin  (A-B)+|  sin  (A-f  B). 

Make  AKC  equal  to  A,  and  BKC,  CKD 
each  equal  to  B;  draw  BE,  CF,  DH,  the 
sines  of  AKB,  AKC,  AKD  ;  join  BD,  and 
through  the  center  K  draw  KNC ;  then 
KN"  is  evidently  the  cosine  of  BKC.  Draw 
also  BML,  NMG  parallel  to  AK,  DH. 
Now,  in  the  similar  triangles  DLB,  NMB, 


K        n         G  F    E  A 

Bince  DB  is  double  of  NB,  DL  (V.  3)  is  double  of  NM ;  to 


PLANE   TRIGONOMETRY. 


189 


DL  add  LII,  BE,  and  to  2NM  add  what  is  equivalent,  2MG ; 
tlien  DH+BE  =  2NG ;  wherefore  NG  is  equal  to  half  the  sum 
of  BE  and  DH,  that  is,  to  i  sin  (A— B)  +^  sin  (A-f  B).  Again, 
in  the  similar  triangles  CFK,  NGK,  we  have  (V.  3,  and  altern- 
ately) CK  :  NK  : :  CF  :  NG;  that  is,  R  :  cos  B  :  :  sin  A  : 
isin  (A— B)+isin(A  +  B). 

Cor.  1.  Hence,  if  R  =  l,  we  have,  by  doubling  the  second 
and  fourth  terms,  and  by  taking  the  products  of  the  extremes 
and  means,  sin  (A— B)4- sin(A  +  B)=;2  cos  B  sin  A;  whence, 
sin  (A+B)=i2  cosB  sin  A— sin  (A— B). 

Co7\  2.  If  B=:A,  the  last  expression  becomes  sirapl}"-  sin  2  A 
=  2  sin  A  cos  A. 


TRIGONOMETRICAL   FORMULA. 

The  lines  hitherto  considered  may  be  computed  for  every 
conceivable  angle,  and  they  will  each  undergo  a  change  of 
value  when  the  angle  passes  through  the  gradations  of  magni- 
tude, hence  they  are  the  functions  of  the  angle,  a  term  imply- 
ing the  connection  between  two  varying  quantities,  that  the 
value  of  the  one  changes  with  the  value  of  the  other,  and  they 
receive  their  values  from  the  ratios  or  proportions  arising  from 
them  and  the  angle.  We  have  considered  the  numerical  values 
only  of  these  functions,  and  the  angles  from  which  they  were 
deduced  were  all  less  than  180  degrees,  which  relate  to  plane 
angles  and  triangles.  And  we  propose  now  to  explain  the 
processes  for  computing  the  unknown  parts  of  rectilinear  trian- 
gles, also  the  nature  and  properties  of  the  angular  functions, 
together  with  the  methods  of  deducing  all  the  formulae  which 
express  relations  between  them. 

When  two  diameters  are  drawn  per- 
pendicularly to  each  other,  they  divide 
the  circle  into  four  equal  parts  called 
quadrants,  which  are  first,  second, 
third,  and  fourth  quadrants,  going  from 
right  to  left,  and  the  functions  have 
certain  algebraic  values  depending 
upon  the  particular  quadrant  in  which 
the  angle  is.  For  instance,  all  the 
lines  estimated  from  AC  upward  are  positive,  and  from  CA 


190  THE   ELEMENTS   OF 

downward  are  negative ;  from  DB  to  the  right  are  positive, 
and  from  DB  to  the  left  are  negative.  In  the  formulae  the 
algebraic  signs  +  and  —  are  used,  the  former  denoting  posi- 
tive, and  the  latter  negative. 

In  the  diagram  it  will  be  seen  that  the  functions  are  all  posi- 
tive in  the  first  quadrant  AEB ;  that  the  sine,  cosecant,  and 
versed  sine  are  positive,  and  the  others  negative  in  the  second 
quadrant ;  that  the  tangent,  cotangent,  and  versed  sine  are 
positive,  and  the  others  negative  in  the  third  quadrant ;  that 
the  cosine,  secant,  and  versed  sine  are  positive,  and  the  others 
negative  in  the  fourth  quadrant.  Hence  we  can  arrange  them 
in  the  following  table  : 

Third  Q.    Fourth  Q. 
—  + 


FmsT  Qttai). 

Second  Q. 

Thir] 

Sine, 

+ 

+ 

Cosine, 

+ 

Tangent, 

+ 

— 

+ 

Cotangent, 

+ 



+ 

Secant, 

+ 

— 

— 

Cosecant, 

+ 

+ 

— 

Versed  sine. 

+ 

+ 

+ 

+ 
+ 

It  is  convenient  to  give  different  signs  to  the  angles  also. 
If  we  suppose  the  angles  to  be  estimated  from  left  to  right, 
they  are  negative,  and  the  sign  of  the  angle  will  afiect  the 
sign  of  its  sine,  but  those  of  its  cosine  remain  the  same. 

From  corollaries  fourth,  fifth,  sixth,  and  seventh  we  can  de- 
duce the  following  equations,  when  A  denotes  the  angle  and 
the  radius  is  unity  : 

Sin  "A + cos 'A =1  (1) 

Sec=A=l  +  tan^A  (2) 

Cosec^'A^l+cot'A  (3) 

r_       .      sin  A 

TanA=: -r  (4) 

cos  A  ^  ' 

^      .      cos  A  ig,\ 

QoiA=-r—.  (5) 

sm  A 


TanAxcotA  =  l  (6) 

1 
cos  A 


Sec  A=-^  (7) 


PLANE   TEIOONOMETRY.  191 

Cosec  A  =  -: T-  (8) 

Sin  A 

Ver,  sin  A=l — cos  A  (9) 

By  the  first  proposition  we  have,  in  a  right-angled  triangle, 
radius  :  cos  of  either  acute  angle  : :  hyp.  :  side  adjacent. 

Hence,  in  the  following  diagram : 

CB:0=CA  cos  C ;  and  DB:0-DA  cosD, 
or  CD=c>CA  cos  C— DA  cos  D. 

Dividing  both  members  of  the  equation 
by  CD,  we  have — 

1=0=— =- cosC— ^=^=-cos  D;  hence  (Trig,  2) 
OD  kjiJ 

CA       sin  D        -  DA       sin  C     ,  sin  D        ^     sin  C 

pr^O- — T   and  7=^^<2>=-^ — r;  hence,  lo-; — r-cosC — : — -r 
CD^^sin  A  CD       sin  A '  sin  A  sm  A 

cos  D,  or  sin  Aosin  D  cos  C— sin  C  cos  D ;  but  the  angle  A 

is  the  difference  of  the  angles  ADB  and  ACB  (I.  20)  ;  hence 

sin  (D— C)=0=sin  D  cos  C — sin  C  cos  D. 

Thus,  from  the  first  and  second  propositions,  by  easy  pro- 
cesses, we  derive  the  formula  for  the  sine  of  the  difference  of 
two  angles,  which  is  expressed  in  the  following  manner  :  The 
sine  of  the  difference  of  any  two  angles  is  equivalent  to  the  sine 
of  the  first  into  the  cosine  of  the  second,  minus  the  cosine  of  the 
first  into  the  sine  of  the  second.* 

The  formula  for  the  sine  of  the  sum  of  two  angles  can  be 
derived  from  the  preceding  by  substituting  the  negative  for 
the  positive  value  of  the  second  angle,  and  bearing  in  mind 
that  in  estimating  an  angle  from  left  to  right,  the  algebraic 
sign  of  its  sine  is  changed,  and  we  get — The  sine  of  the  sum 
of  any  two  angles  is  equivalent  to  the  sine  of  the  first  into  the 
cosine  of  the  second,  plus  the  cosine  of  the  first  into  the  sine  of 
the  second. 

The  formula  for  the  cosine  of  the  sum  of  two  angles  can  be 
derived  from  the  preceding,  by  substituting  the  trigonometrical 
values  of  the  functions  when  the  sum  of  the  angles  is  greater 
than  a  right  angle,  and  remembering  that  the  sine  of  an  angle 
becomes  the  cosine  of  its  complement,  and  we  get — The  cosine 

*  The  pupil  would  be  much  instructed  by  converting  this  and  the  fol- 
lowing expressions  into  their  equivalent  algebraic  formula;. 


192  THE   ELEMENTS    OF 

of  the  sum  of  two  angles  is  equivalent  to  the  cosine  of  the  first 
into  the  cosine  of  the  second,  minus  the  sine  of  the  first  into  the 
sifie  of  the  second. 

By  similar  substitutions  in  the  second  formula,  we  get  the 
formula  for  the  cosine  of  the  difference  of  two  angles,  expressed 
as  follows:  The  cosine  of  the  difference  of  two  angles  is  equiva- 
le?it  to  the  cosine  of  the  first  into  the  cosine  of  the  second,  ^>^ws 
the  sine  of  the  first  into  the  sine  of  the  second. 

The  other  corresponding  formuliB  are  obtained  by  substitut- 
ing the  respective  equations  for  the  trigonometrical  fmictions 
derived  from  the  fourth,  fifth,  sixth,  and  seventh  corollaries  for 
the  values  of  the  various  functions ;  for  instance,  to  derive  the 
formula  for  the  tangent  of  the  sum  of  two  angles,  we  substi- 

tute,  tanA  = r  in  the  second  and  third  formulse,  getting 

'  cos  A 

,,  „,  sin  (A+B)  sin  A  cos  B  4- cos  A  sin  B  „  ,  ^^ 
t""  (^+^)=co4At-B)  =  co.AcosB-sinAshrB'  ""^  '"^ 
ducing,  we  have — The  tangent  of  the  sum  of  two  angles  is 
equicalent  to  the  tangent  of  the  first, plus  the  tangent  of  the  sec- 
ond, divided  hy  the  square  of  the  radius,  minus  the  tangent  of 
the  first  into  the  tangent  of  the  second.  And  in  a  similar  way 
we  get  the  other  formulae  for  the  various  functions. 

When  the  angle  and  radius  are  known,  we  can  get  the  form- 
ula for  the  sine  of  double  the  angle  by  making  the  two  angles 
equal  in  the  second  formula,  and  we  have — The  sine  of  twice 
an  angle  is  equivalent  to  twice  the  sine  of  the  angle  into  the 
cosine  of  the  angle. 

In  similar  manner  we  derive  the  formula  for  the  cosine  of 
double  the  angle,  and  substituting  the  equation  sin''A=l — 
cos'  A  in  the  third  formula,  we  get — The  cosine  of  twice  an 
angle  is  equivalent  to  twice  the  square  of  the  cosine  of  the  an- 
gle, minus  the  square  of  the  radius. 

Thus  by  substitution  of  the  several  equations  in  the  respec- 
tive formulae,  we  can  derive  the  other  functions  of  double  the 
ano-le  when  the  ansfle  and  radius  are  known. 

From  the  formula  for  the  cosine  of  double  an  angle,  we  can 
get  by  substitutions  and  reductions  the  formula  for  the  sine  of 
half  an  angle,  and  expressed  as  follows — The  sine  of  half  an 
angle  is  equivalent  to  the  square  root  of  half  the  difference  of 


PLANE   TRIGONOMETRY.  193 

the  radius  and  cosine  of  the  angle.  In  a  similar  way  we  ob- 
tain— The  cosine  of  half  an  angle  is  equivalent  to  the  sqf/are 
root  of  half  the  sum,  of  the  radius  and  cosine  of  the  angle  ^ 
and — The  tangent  of  half  an  angle  is  equivalent  to  the  sine  of 
the  angle  divided  hy  the  sum  of  the  radius  and  cosine  of  the 
angle  ;  and  so  on  for  the  other  functions. 

By  adding  and  subtracting  the  various  formulae  already  men- 
tioned, we  obtain  a  great  number  of  consequences  which  are 
useful ;  it  will  suffice  to  consider  a  few  of  them.  From  the 
first  four  we  obtain — The  sine  of  the  sum  of  two  angles  added 
to  the  sine  of  the  difference  of  the  same  angles  is  equivalent  to 
twice  the  sine  of  the  first  into  the  cosine  of  the  second.  The 
siiie  of  the  sum  of  two  angles  diminished  by  the  sine  of  the 
difference  of  the  same  angles  is  equivalent  to  txcice  the  sine  of 
the  second  into  the  cosine  of  the  first.  TJie  cosine  of  the  sum, 
of  two  angles  increased  by  the  cosine  of  the  difference  of  the 
same  angles  is  equivalent  to  twice  the  cos  ne  of  the  first  into 
the  cosine  of  the  second.  The  cosine  of  the  difference  of  two 
angles  dimhiished  by  the  cosine  of  the  sum  of  the  same  angles 
is  equioalent  to  twice  the  sine  of  the  first  into  the  sine  of  the 
second.  These  are  very  useful,  because  they  change  the  pro- 
ducts of  sines,  cosines,  and  other  functions  from  superficial  into 
linear  sines,  cosines,  etc. 

By  the  substitution  of  algebraic  symbols  into  the  preceding 
equations,  we  obtain  certain  analogies  which  give  algebraic  ex- 
pressions to  so  many  theorems,  as  follows : 

Sum  of  sines  :  Dif.  of  sines  : :  tan  of  half  Sum  :  tan  of  half  Dif. 

Sum  of  sines  :  Sum  of  cos    ::  tan  of  half  Sum  :  radius. 

Sum  of  sines  :  Dif  of  cos     ::  cot  of  half  Dif  :  radius. 

Dif  of  sines  :  Sum  of  cos   : :  tan  of  half  Dif  :  radius. 

Dif  of  sines  :  Dif  of  cos    ::  cot  of  half  Sum  :  radius. 

Sum  of  cos    :  Dif  of  cos     ::  cot  of  half  Sum  :  tan  of  half  Dif. 

Sum  of  sines  :  sine  of  Sum  : :  cos  of  half  Dif    :  cos  of  half  Sum. 

Dif  of  sines  :  sine  of  Sum  ::  Sine  of  half  Dif  :  sine  of  half  Sum. 

INYESTIGATIONS    OF   THE    METHODS    OP    COMPUTING   TABLES    OF 
SINES,    TANGENTS,    AND    SECANTS. 

From  what  has  been  shown  in  relation  to  the  previous  form- 
12 


194r  THE   ELEMENTS   OF 

nlae,  it  will  be  noticed  that  they  all  proceed  from  the  first — and 
we  derive  the  ^rst  from  the  first  and  second  propositions, 
namely,  from  the  analogies  that  in  a  right-angled  triangle : 
hypothennse  :  radius  : :  one  of  the  legs  :  sine  of  opposite  an- 
gle ;  hence,  when  we  know  the  length  of  the  hypothenuse,  leg, 
and  radius,  we  can  determine  the  sine  of  the  angle  opposite  the 
leg.  Then  (Trig,  def  V,  cor.  4)  cos"A=l— sin'A ;  that  is,  the 
square  of  the  radius,  minus  the  square  of  the  sine  of  the  angle, 
IS  equivalent  to  the  square  of  the  cosine  of  the  angle;  then, 
square  root  of  the  difierence  between  the  squares  of  the  radius 
and  sine  is  equivalent  to  the  cosine  of  the  angle;  and  the  other 
functions  are  derived  from  the  equations  resulting  from  the 
fourth,  fifth,  sixth,  and  seventh  corollaries  of  the  definitions. 

Now,  the  chord  of  60  degrees  (III.  25,  cor.  4)  is  equal  to  ra- 
dius, and  when  radius  is  unity,  the  cosine  of  30°  is  0.5  ;  hence, 
sine  of  30°=  ^/(l — cos^30°)=:  4/.75,  Bisecting  this  angle  we 
get  from  the  formula  of  cosine  of  half  an  angle,  cos  15°= 
■iVl+cos  30°,  and  seventeen  such  bisections  give  cos  1'^= 
.999999299 ;  hence,  sin  V  can  be  obtained. 

Some  mathematicians  divide  3.14159265358979,  etc.,  into  as 
many  equal  parts  as  there  are  seconds  or  minutes  in  180°,  and 
express  the  value  of  the  sine  of  one  second  or  minute  by  one  of 
these  equal  parts,  contending  that  the  sine,  chord,  or  arc  of  so 
small  an  angle  differ  very  imperceptibly  from  each  other; 
when  this  quantity  is  used,  the  cosine  of  one  second  becomes 
.999999957.  It  will  thus  be  seen  that  the  cosines  obtained  by 
these  methods  differ  very  little  from  each  other,  and  for  ordi- 
nary purposes  either  will  give  results  sufiiciently  accui'ate ;  but 
when  the  greatest  exactness  is  desired,  the  first  method  should 
be  used,  because  the  sine  of  an  angle  is  a  straight  line  and  can 
never  coincide  with  the  arc  which  measures  (I.  def  19)  the  an- 
gle, however  so  small  the  angle  be  reduced,  and  (V.  25,  scho.) 
3.1415926,  etc.,  is  the  approximate  relation  of  the  diameter  and 
circumference.  Hence,  the  first  method  is  a  pure  deduction 
from  geometrical  and  trigonometrical  principles.  When  the 
angle  is  a  right  angle,  the  sine  and  cosine  of  the  angle 
are  equal,  and  the  equations  of  the  tangent  and  cotangent 
will  give  the  values  of  those  functions;  and  when  the  angle  ex- 
ceed a  right  angle,  the  formula  for  the  tangent  of  the  sum  of 


PLANE   TRIGONOMETKY. 


195 


two  angles  can  be  used,  making  the  right  angle  one  of  the 
angles  and  the  excess  the  other  angle. 

When  the  converging  series  are  used,  any  angle  less  than 
90''  is  expressed  by  cc,  and  the  formulae  for  the  functions  are: 

Sm  a;=a; 1 -4-  etc. 

1.2.3^1.2.3.4.5       1.2.3.4.5.6.7^ 

x'  X*  a;*        .     ^ 

Cos  x=:l 1 \-  etc. 

1.2^1.2.,3.4        1.2.3.4.5.6^ 

^  ,    a'     2af      17a;'        62a;' 

Tan  x=zx-\ 1 1 — - — -+-^ — --:+  etc. 

.  3  ^3.5^3=.5.7      3'.5.7.9^ 

Cotcc^----  — -^^-  — -etc. 

,  ,   a;'  5a;*  61a;* 

Sec  a;=lH — --\ 1 \-  etc. 

1.2^1.2.3.4^1.2.3.4.5.6 

1         X  1x^  S\x' 

Cosec  x= 1 1 1 4-  etc. 

a;  ^1.2.3^3.4.5.6     3.4.5.6" 

Now,  when  the  base  of  the  Napierian  logarithms  is  used,  €= 
2.7182818,  and  the  following  formulie  will  give  the  sine  and  co- 
sine of  any  angle  x,  from  which  the  other  functions  can  be 
obtained : 

Sm  a;= -^ ;  and  cosa;  = 


2  ^^  2 

Having  obtained  the  sine  and  cosine  of  any  angle  by  either 
of  the  foregoing  formulae,  we  can  get  the  sine  of  twice  the  an- 
gle by  the  consequences  fi-om  adding  and  subtracting  the  first 
four  formulae,  page  193,  from  which  we  derive — 

2  cos  a;  X  sin  x  —  sin  0  =  sin  2  ar, 

2  cos  X  X  sin  2  a;  —  sin  a;  =  sin  3  x, 

2  cos  XX  sin  3  a;  —  sin  2  a;  =  sin  4  a-, 
etc.,        etc.,         etc.,^      etc. 

Or  by  multiplying  the  first  two  formulae,  page  191,  and  sub- 
stituting the  value  of  the  square  of  the  cosine,  we  can  deter- 
mine new  formulae  for  further  computation  after  having  found 
the  sines  of  x  and  2  x. 

Sin  (a;+2a;)  sin  (x— 2  x)=sin'a;— sin'2r ;  hence,  sin  (a;+2a-) 


196  THE    ELKMENT8    OF 

sin  (.c— 2  3-)  =  (sin  x-\-s\n  2  x)  (sin  a:— sin  2  a;);  or,  sin  (x — 2a'): 
sin  X — sin  2x  ::  sin  x+sin  2x  :  sin  {x-{-2  x)  ;  applying  lliia 
proportion,  we  have, 

Sin  X  :  sin  2  a;— sin  x  : :  sin  2  x  +  sin  x  :  sin  3  x. 
Sin  2  X  :  sin  3  J— sin  x  : :  sin  3a:  +  sin  a:  :  sin  4  x. 
Sin  3  a;  :  sin  4  x — sin  a;  : :  sin  4  .r+sin  a;  :  sin  5  x. 
etc.,  etc.,  etc.,  etc. 

These  last  formulje  will  give  the  natural  functions  of  the  an- 
gles, but  to  avoid  the  operations  of  multiplication  and  divi- 
sion, and  employ  the  simpler  operations  of  addition  and  sub- 
traction, tables  are  constructed  giving  the  logarithmic  values 
of  the  several  functions  of  the  ansrles. 

As  the  sine  and  cosine  of  an  anorle  are  the  leers  of  a  rieht- 
angled  triangle,  and  the  hypothenuse  is  the  radius  of  the  arc 
which  measures  the  angle,  for  the  convenience  of  logarithms 
the  hypothenuse  or  radius  is  considered  as  10,000,000,000,  and 
its  logarithm  is  10. 

The  sines,  cosines,  tangents,  and  cotangents  are  the  only- 
functions  put  in  the  tables,  as  the  other  functions  are  easily 
found  from  them. 

TRIGONOMETRICAL   PROBLEMS. 

The  principles  which  have  been  thus  established,  enable  us  to 
solve  all  the  elementary  cases  of  plane  trigonometry.  Now,  of 
the  three  sides  and  three  angles  of  a  triangle,  some  three,  and 
those  not  the  three  angles,  must  be  given  to  determine  the  tri- 
angle (I,  14,  scho.),  and  the  resolution  of  plane  triangles  may 
therefore  be  reduced  to  the  three  followinsr  cases : 

I.  When  a  side  and  the  opposite  angle,  and  either  another 
side  or  another  angle  are  given  ; 

II.  When  two  sides  and  the  contained  anijle  are  aiven ; 

III.  When  the  three  sides  are  c^iven 


o' 


The  FIRST  CASE  is  solved  on  the  principle  (Trig.  2)  that  the 
sides  are  proportional  to  the  sines  of  the  opposite  angles.  Thus, 
if  A,  B,  a  be  given,  add  A  and  B  together,  and  take  the  sum 
from  180°;  the  remainder  (I.  20)  is  C.     Then  b  and  c  will  be 


PLANE    TRIGONOMETRY.  197 

found  by  the  following  analogies  :  sin  A  :  sinB  :  :  «  :  Jy  and 
sin  A  :  sin  G  :  :  a  :  c. 

If,  again,  a,  b,  A  be  given,  we  compute  B  "by  the  analogy, 
a  :  6  :  :  sin  A  :  sin  B.  Then,  C  is  found  by  subtracting  the 
sura  of  A  and  B  from  180°,  and  c  by  the  analogy,  sin  A  :  sin  C 

When  in  this  case  two  unequal  sides,  and  the  angle  opposite 
to  the  less,  are  given,  the  angle  opposite  to  the  greater  (Tkig. 
defs,  cor.  8)  may  be  either  that  which  is  found  in  the  table  of 
Bines  or  its  supplement ;  and  thus  the  problem  admits  of  two 
solutions  (I.  3,  case  4). 

If  in  this  case  one  of  the  anHes  be  a  riuht  anole,  the  solution 
is  rather  easier;  as,  by  the  second  corollary  to  the  definitions, 
the  sine  of  that  angle  is  equal  to  the  radius.  The  same  con- 
clusion may  also  be  obtained  by  means  of  the  first  propo- 
sition. 

To  exemplify  the  solution  of  this  case,*  let  az=l3  yards,  5= 
15  yards,  and  Ar=53°  8', — to  resolve  the  triangle;  and  the 
operation  by  means  of  logarithms  will  be  as  follows: 


As  a 

13 

1.113943 

:  b 

15 

1.176091 

: :  sin  A 

67° 

8^ 
23^ 

9.903108 

:  sinB 

9.965256 

or 

112'=' 

37' 

As  sin  A 

9.903108 

:  sin  C 

59» 

29' 

9.935246 

: :       a 

14 

1.113943 

:        c 

1.146081 

As  sin  A 

9.903108 

;  sin  C 

14" 

15' 

9.391206 

::        a 

4 

1.113943 

:         c 

0.602041 

*  For  logarithmic  computations  the  pupil  is  referred  to  the  Tables 
now  in  preparation  by  Prof.  Docharty,  of  the  College  of  the  City  of  New 
York,  or  the  Tables  computed  by  Prof  Davies,  of  tlie  United  States 


198  THE   ELEMENTS    OF 

In  these  operations,  to  find  the  fourth  term,  the  second  and 
third  terms  are  added  together,  and  the  first  is  taken  from  the 
sum.  This  may  be  done  very  easily,  in  a  single  operation,  by 
addins:  the  fisrures  of  the  second  and  third  terms  successively 
to  what  remains  after  taking  the  right-hand  figure  of  the  first 
term  from  10,  and  each  of  the  rest  from  9,  and  rejecting  10 
from  the  final  result.  Thus,  in  the  first  operation,  we  have  8 
and  1  are  9,  and  7  are  16;  then  1  and  9  are  10,  and  5  are  15, 
etc.  It  is  still  easier,  however,  when  the  quantity  to  be  sub- 
tracted is  a  sine,  to  use  the  cosecant,  and  when  it  is  a  cosine, 
to  use  the  secant,  each  diminished  by  10,  and  then  to  add  all 
the  terms  together.  The  reason  of  this  is  evident  from  the  na- 
ture of  logarithms,  and  from  the  fifth  corollary  to  the  defini- 
tions of  Trigonometry.  In  like  manner,  when  the  number  to  be 
subtracted  is  a  tangent,  or  cotangent,  we  may  use  in  the  former 
case  a  cotangent — in  the  latter,  a  tangent,  subtracting  in  each 
case  10,  either  at  first  or  afterward. 

This  example  evidently  belongs  to  the  doubtful  case;  and 
hence  we  have  two  values  for  each  of  the  quantities  B,  C,  and 
c ;  and  therefore  two  analogies  are  requisite  for  finding  the 
values  of  c. 

The  SECOJTD  CASE  is  solved  by  means  of  the  third  and  fourth 
propositions.  Thus,  if  a,  h,  C  be  taken,  take  C  from  180°,  and 
(I.  20)  the  remainder  is  A  4-B.  Take  the  half  of  this,  and  then, 
by  the  third  proposition,  as  a-\-h  :  a—h  :  :  tan|  (A  +  B)  :  tan 
\  (A— B).  This  analogy  gives  half  the  difference  of  A  and  B  ; 
and  (II.  12,  scho.)  by  adding  this  and  i  (A  +  B)  together,  A, 
the  greater  angle,  is  obtained,  while  B  is  found  by  taking  \  (A 
— B)  from-|(A+B).  The  remaining  side  will  be  calculated 
(Tkig.  4)  by  means  of  either  of  the  following  analogies,  and 
"by  employing  both,  an  easy  verification  of  the  process  is  ob- 
tained ; 

as  cos  i  (A— B)  :  cos  ^  (A  +  B)  ::a\h:c;  and 
sin  \  (A— B)  :  sin  |  (A  +  B)  wa-h-.c. 

When  the  given  angle  C  is  a  right  angle,  the  solution  is 
most  easily  efiected  by  means  of  the  first  proposition  of  Trig- 
Military  Academy,  or  those  of  Frof.  Loomis,  of  Yale  College,  New 
llaven,  Connecticut. 


PLANE    TRIGONOMETRY.  199 

onometry;  the  oblique  angles  being  obtained  by  the  analogy, 
a  :  b  ::  U  :  tan  B,  or  cot  A ;  and  the  hypotheinise  either  by 

c. 


the  analogy,  R  :  sec  B  :  :  a  : 

c,  or  sin  A  :  R  :  :  a  : 

As  an  example, 

As  a-\-b 

99.98           1.999913 

:  a — b 

14.V8           1.169G74 

:  :  tan  ^  (A+B) 

61°  37'i   10.267498 

:tan^(A— B)     15°  18'i     9.437259 
Hence  A  =  76°  56',  and  Bi=46o  19' 

As  cos  i  (A— B)     15°   18'i      9.984311 

:  cos  i  (A+B)  9.676913 

:  :  a+b  1.999913 


C  49.20  1.692515 

As  sin  I  (A— B)  9.421626 

:  sin  i  (A  +  B)  9.944411 

::  a-b  1.169674 


:  c        49.26  1.692459 

Half  the  difference  of  A  and  B  is  here  taken  as  15°  18'^. 
When  determined  accurately,  however,  it  is  found  to  be  15° 
18'  23",  Hence  the  cause  of  the  slight  difference  in  the  loga- 
rithm of  c,  as  obtained  by  the  two  different  analogies.  It  is 
plain  that  after  A  and  B  are  computed,  c  might  also  be  found 
by  means  of  the  first  case,  by  either  of  the  analogies;  sin  A  : 
Bin  C  :  :  a  :  c,  and  sin  B  :  sin  C  :  :  b  :  c.  The  foregoing 
method,  however,  is  much  preferable. 

The  THIRD  CASE  may  be  solved  by  means  of  the  fifth,  sixth, 
or  seventh  proposition.  Thus,  assuming  a  (see  the  figures  for 
the  fifth  proposition)  as  base,  we  have  a  to  b-\-c  as  b—c,OT- 
c—b  to  a  fourth  proportional.  If  this  be  less  than  BC,  it  is  the 
difference  of  the  segments  BD,  DC,  in  the  first  figure  ;  and  if 
half  of  it  and  half  of  the  base  be  added  together,  the  sum  will 
be  the  greater  segment,  while  the  less  will  be  found  by  taking 
half  that  proportional  from  half  the  base.     If  the  fourth  pro- 


200  THE    ELEMENTS   OF 

portional  be  greater  than  the  base,  it  is  the  sura  of  the  seg- 
ments in  the  second  figure,  and,  as  before,  the  segments  are  the 
sum  and  difference  of  half  the  proportional  and  half  the  base. 
Then,  by  resolving,  by  the  first  case,  the  two  riglit-angled  tri- 
angles ADB,  ADC,  in  which  there  are  given  the  hypothenuses 
AB,  AC,  and  the  legs  BD,  CD,  the  angles  B  and  ACD  will  be 
obtained,  which,  in  the  first  figure,  are  two  of  the  required  an- 
gles ;  while  in  the  second,  the  angle  C  is  the  supplement  of 
ACD. 

Again  :  by  adding  the  three  sides  together,  and  talcing  half 
the  sum,  the  value  of  5  is  obtained;  and,  in  applying  the  sixth 
proposition,  the  sides  containing  the  required  angle  are  to  be 
separately  taken  from  sy  but,  in  applying  the  seventh,  only 
the  side  opposite  to  the  required  angle  is  to  be  subtracted; 
while  if  all  the  three  sides  be  subtracted  successively,  another 
mode  of  solution  is  furnished  by  the  second  and  third  corolla- 
ries to  the  seventh  proposition.  This  last  method  is  preferable 
to  any  of  the  others,  when  it  is  necessary  to  determine  all  the 
angles ;  and  if  they  be  all  computed  by  means  of  it,  the  cor- 
rectness of  the  operation  is  ascertained  by  trying  whether  their 
sum  is  180°. 

To  exemplify  the  last  of  these  methods,  let  a=Qld,  b—53ly 
and  c=429  ;  to  compute  the  angles. 

Here,  by  adding  the  three  sides  together,  we  obtain  1645, 
the  half  of  which,  822.5,  is  s.  Then,  by  taking  from  the  three 
sides  successively,  we  find  s  —  a= 143.5,  s  —  &= 285.5,  and  s — 
c=393.5.  The  rest  of  the  operation,  the  subtraction  in  the  first 
part  of  which  may  be  performed  in  the  manner  pointed  out  in 
the  example  for  the  first  case,  is  as  follows : 


8 

8 — a 

822.5 
143.5 

2.156852  ) 

s      h 

285.5 

2.455606 

8 C 

393.5 

2.594945 

2) 

19.978563 

n  ^A  44° 

17'i 

9.989281 

A=88° 

35' 

PLANE   TRIGONOMETRY. 

tan  i  A  9.989281  )  ^^^ 

log{s  —  a)  2.156852) 

12.146133  )      ,  . 
c  subt. 
log{s  —  b)  2.455606) 


201 


tan  IB  26°  V'i  9.690527 

B  =  520  15' 

12.146133)^^^^ 
log  {s  —  c)  2.594945  ) 


tan  i  C  19°  35'  9.551188 

C  =  39°  10' 

In  the  first  part  of  this  operation,  the  halving  of  the  loga- 
rithm serves  for  the  extraction  of  the  square  root.  The  remain- 
der of  the  work  consists  in  adding  together  tan  ^  A  and  log 
{s  — a),  and  subtractingAog  {s —  b)  and  log  (s  —  c)  successively 
from  the  sum.  This  method  of  solution  is  remarkably  easy, 
requiring  for  the  entire  operation  only  four  logarithms  to  be 
taken  from  the  tables  ;  and  affording  at  the  same  time  a 
most  satisfactory  verification  by  the  addition  of  the  three 
angles,  when  found.  The  preparatory  part  of  the  process 
also  admits  of  an  easy  verification,  as  the  sum  of  the  three 
remainders  s  —  «,  s  —  b,  s  —  c  is  equal  to  the  half  sum. 

Prob.  I. — Let  it  be  required  to  find  the  height  of  an  accessi- 
ble object  AB,  standing  on  a  horizontal  plane. 

On  the  horizontal  plane  take  a  station  C,  and  measure  with 
a  line,  a  chain,  or  any  such  instrument,  the  dis- 
tance CB  to  the  base  of  the  object ;  and  with  a 
quadrant,  a  theodolite,  or  other  angular  in  tru- 
ment,  measure  the  angle  BCA,  called  the  angle 
of  elevation.  Then,  since  B  is  a  right  angle,  the 
height  AB  will  be  found  (Trig.  1)  by  the  analo- 
gy,^R  :  tan  C  :  :  CB  :  BA. 

This  gives  the  height  of  A  above  CB,  the  horizontal  line 
passing  through  the  eye  of  the  observer ;  and  tlierefore  to  find 
the  entire  height,  AB  must  be  increased  by  the  height  of  hia 


202  THE   ELEMENTS   OF 

eye  above  the  base  or  the  object.  The  like  addition  must  be 
made  in  every  problem  of  this  kind,  when  the  angle  of  eleva- 
tion above  the  horizontal  line  is  given. 

pROB.  n. — To  find  the  height  of  an  object  AB,  stanaing  on 
a  horizontal  plane,  but  inaccessible  ow  account  of  the  uneven- 
ness  of  the  ground  near  its  base,  or  the  intervefition  of  obsta- 
cles. 

In  a  straight  line  passing  through  the  base  of  the  object  take 
two  stations  C,  D ;  and  measure  CD,  and 
the  two  angles  of  elevation  BCA,  BDA. 
Then  (I.  20)  CAD  is  the  difference  of  ACB, 
ADB ;  and  (TFao.  2)   sin  CAD  :  sin  D  : : 
DC  :  CA.     Again  (Trig.  1),  R  :  sin  ACB 
:  :  AC  :  AB  ;  whence  AB  will  be  found. 
The  computation  will  be  rendered  rather  more  easy  by  mul- 
tiplying together  (IV.  15)  the  terms  of  the  two  analogies,  and 
dividing  the  third  and  fourth  terms  li^'the  result  by  AC ;  as  by 
this  means  we  get  the  analogy  li  X  sin  CAD  :  sin  D  x  sin  BCA 
:  :  DC  :  AB.     Hence,  to  find  the  logarithm  of  AB,  to  the  loga- 
rithm of  DC  add  the  logarithmic  sines  of  D  and  BCA,  and  from 
the  sum  take  the  sine  of  CAD  and  the  radius. 

Prob.  Ill, —  To  find  the  distance  oftxno  objects  A  a7idB  on  a 
horizontal  plane. 

This  may  be  effected  in  different  ways  according  to  circum- 
stances. 

1.  A  base  AC  may  be  taken,  terminated  at  one  of  the  ob- 
jects. The  angles  A  and  C,  and  the  side  AC 
are  then  measured  ;  and  the  required  distance 
AB  is  found  by  the  analogy,  sinB  :  sin  C  :  : 
AC  :  AB. 

2.  This  method  fails  if  the  objects  A  and  B 
be  not  visible  from  one  another,  as  then  the 
angle  A  can  not  be  measured.  In  this  case,  a 
station  C  may  be  taken  as  before,  from  which  both  A  and  B 
are  visible.  Then,  having  measured  the  angle  C,  and  the  sides 
AC,  BC,  we  compute  the  distance  AB  by  means  of  the  second 
case  of  trigonometry. 


PLANE   TKIGONOMETKY. 


203 


3.  When  from  inequalities  in  the  ground,  or  other  causes,  the 
preceding  methods  are  inapplicable,  the  solution  may  be  effect- 
ed in  the  following  manner:  Measure  a  base  CD,  such  that  A 
and  B  are  both  visible  from  each  of  its  extremities ;  measure, 
also,  the  two  angles  at  C,  and  the  two  at  D.  Then,  by  the 
first  case  in  trigonometry,  -we  compute  AC  in  the  triangle 
ACD,  and  BC  in  the  triangle  BCD  ;  from  which,  and  from  the 
contained  angle  ACB,  AB  is  computed  by  means  of  the  second 
case.  The  operation  may  be  verified  by  computing  AD,  BD, 
by  the  first  case,  and  thence  AB  by  the  second  case. 

Prob.  IV. — Let  AFB  be  a  great  circle  of  the  earth,  supposed 
to  be  a  sphere;  E  a  point  in  the  diameter  B A  produced,  EF  a 
straight  line  touching  the  circle  in  F,  and  ED  a  straight  line  in 
its  plane,  per2Jendicular  to  AB;  it  is  required  to  compute  the 
angle  DEF,  and  the  straight  line  EF. 

Draw  the  radius  CF.  Then,  since  (III.  12)  CFE  is  a  right 
angle,  we  have  (hyp.  and  I.  20,  cor.  3) 
DEC=CEF+ECF."  Take  away  CEF, 
and  there  remains  DEF=:ECF.  Now 
(III.  21)  EF-  =  BE.EA.  Hence  EF  will 
be  found  by  adding  AE  to  AB,  multiply- 
ing the  sum  by  EA,  and  extracting  the 
square  root.     To  find  CE,  add  AE  to  the 

radius  AC.     Then  (Trig.  1)   CE  :  EF  :  :  

R  :  sin  ECF,  or  sin  DEF ;  or  CE  :  CF  : :  ~ 

Pv  :  cos  ECF,  or  cos  DEF. 

8cho.  These  examples  have  been  selected  from  the  "  Ele- 
ments of  Plane  Trigonometry,"  by  Prof  Thomson,  of  the  Uni- 
versity of  Glasgow,  since  they  exhibit  in  the  simplest  manner 
the  elementary  principles  of  trigonometrical  computation. 


4* 


THE  ELEMENTS  OF  SPHERICAL  TRIGONOMETRY. 

DEFINITIONS. 

1.  Spherical  Trigonometry  explains  the  processes  of  cal- 
culating the  unknown  parts  of  a  spherical  triangle  when  any 
three  parts  are  given;  and  certain  formulae  derived  from  Plane 
Trigonometry  are  employed  to  express  the  relations  between 
the  six  parts  of  a  spherical  triangle. 

2.  A  spherical  triangle  is  that  portion  of  the  surface  of  a 
sphere  bounded  or  contained  by  the  arcs  of  three  great  circles 
intersecting  each  other ;  the  spherical  triangle  being  formed  by 
three  planes  passing  through  the  sphere,  and  intersecting  each 
other,  each  angle  of  the  triangle  (VI.  24)  is  contained  by 
the  tangents  of  the  sides  at  their  point  of  intersection,  and  is 
measured  by  the  arcs  of  great  circles  described  from  the  ver- 
tices as  poles,  and  limited  by  the  sides  of  the  triangle  produced 
if  necessary.  Also,  the  angles  of  a  spherical  triangle  vary 
(VL  24,  cor.  2)  between  two  and  six  right  angles. 

3.  The  spherical  angle  being  contained  by  the  tangents  of 
the  sides  at  their  point  of  intersection,  the  properties  of  the 
spherical  triangle  are  explained  by  means  of  Plane  Trigonome- 
try, and  its  analogies  are  applied  to  imaginary  rectilineal  tri- 
angles, the  sides  of  the  spherical  triangle  being  regarded 
functions  of  rectilineal  angles  having  the  sides  of  the  spherical 
triangle  as  arcs  measuring  (I.  def  19)  them.  Spherical  Trig- 
onometry treats  of  the  angles  at  the  apex  of  a  triangular  pyra- 
mid; but  Plane  Trigonometry  treats  oi plane  angles  ^  therefore 
Spherical  Trigonometry  treats  of  solid  angles. 

4.  Let  ABC  be  a  spherical  triangle,  and  H  the  center  of  the 
sphere ;  the  angles  of  the  tnangle  are  equal  to  the  angles  in- 
cluded by  the  planes  IIAB,  HAC,  and  IIBC  (VI  24),"which 
are  the  angles  formed  by  the  planes  at  the  apex  of  a  triangular 
pyramid,  and  the  arcs  AB,  BC,  and  CA  measure  the  angles  on 
the  planes  at  the  apex  of  the  pyramid  AHB,  BHC,  AHC  re- 


SPHERICAL   TRIGONOMETRY. 


205 


spcctively.  And  we  can  represent  the  side  opposite  the  angle 
A  by  a,  the  side  opposite  the 
angle  B  by  b,  and  the  side  op- 
po>ite  the  angle  C  by  c.  On 
the  line  HA  take  any  point  as 
L,  and  draw  perpendiculars  as 
FL,  LG  to  HA.  Then,  GLF 
will  be  equal  to  the  angle  A, 
LEG  equal  to  B,  and  FGL 
equal  to  C  ;  and  the  sides  CB, 
BA,  and  AC  of  the  spherical  triangle  ABC  will  measure  the 
angles  CHB,  AHB,  and  AHC  respectively ;  hence  these  an- 
gles are  denoted  by  a,  c,  b. 

5.  If  FG  be  joined  in  the  triangles  FHG  and  FLG,  we  will 
have  (Plane  Trig.,  6  cor.), 

HF'+HG'-FG* 


cos  BHC=cos  a= 


cos  FLG = cos  A= 


2HFxHG 
LF=+LG=-FG* 


2LGxLF      • 

Reducing   and   subti-acting   second   from  the  first,  we  will 
have, 

2  [cos  a  (HFxHG)— cosA  (LGxLF)]=.2HLl 

Dividing  both  members  by  2  (HFxHG),  we  get. 


cos  a— cos  A 


LGxLF      HL    HL 


HFxHG" 


HF^HG" 


Since  regular  and  similar  polygons  have  their  perimeters 
proportionate  to  their  apothems,  and  circles  have  their  circum- 
ferences proportionate  to  their  diameters  (V.  14,  cor.  3),  the 
sine  of  an  angle  is  the  ratio  of  the  radius,  or  the  hypothenuse 
of  a  liglit-angled  triangle  to  the  perpendicular  from  the  vertex 
of  the  right  angle  to  the  hypothenuse. 

„  LG      ,    ,  LF       .       HL  HL 

Hence  we  get  ^-  =  sin  5,  g^ = sm  e,  g-, = cos  c,  jj^ = cos  b. 

Substituting  and  ti-ansposing,  we  derive  the  formula, 

cos  a=cos  b  cos  c+sin  b  sin  c  cos  A,  ) 
cos  b  =  cos.  a  cos  c+sin  a  sin  c  cos  B,  /•    (1) 
cos  c=cos  a  cos  6+  sin  a  sin  b  cos  C.  ) 


206 


THE   ELEMENTS   OF 


Or,  The  cosine  of  either  side  of  a  spherical  triangle  is  equal 
to  the  product  of  the  cosiyies  of  the  other  two  sides  increased  by 
the  product  of  their  sines  into  the  cosine  of  the  an^le  included 
hy  them. 

The  three  formulae  show  the  relations  between  the  six  parts 
of  a  spherical  triangle  such,  that  if  any  three  of  them  be  given, 
the  other  three  can  be  determined. 

6.  Then  (VI.  24,  cor.  1),  if  we  denote  the  angles  and  sides  of 
the  spherical  triangle  polar  to  ABC,  respectively,  by  A',  B'',  C^, 
a\  h\  c',  we  will  have, 

«'=180°— A,  d'=180°— B,  c'=180O— C, 
A'==180°— a,  B'r=l80°— 6,  C"=180O— c. 
Since  any  of  the  formulae  (l)  is  applicable  to  polar  spherical 
triangles,  we   have,  after  substituting    the   respective   values 
and  changing  the  signs  of  the  terms,  other  formulae  : 

cos  A=sin  B  sin  C  cos  a  —  cos  B  cos  C,  \ 
cos  B=sin  A  sin  C  cos  h  —  cos  A  cos  C,  \    (2) 
cos  C=sin  A  pin  B  cos  c  —  cos  A  cosB.  ; 

Or,  The  cosine  of  either  angle  of  a  spherical  triangle  is  equal 
to  the  product  of  the  sines  of  the  two  other  angles  into  the  co- 
sine of  their  included  side,  diminished  by  the  product  of  the 
cosines  of  those  angles. 

V.  Transposing  the  first  and  second  formulae  (l)  we  get, 
cos  a  —  cos  h  cos  c=sin  h  sin  c  cos  A,^ 
cos  b  —  cos  a  cos  c=:sin  a  sin  c  cos  B ; 

respectively  adding  and  subtracting,  we  get, 

cos  a+ cos  6— cos  c  (cos  a+ cos  5)=sin  c  (sin  5  cos  A+ sin  a 

cos  B), 
cos  a — cos  &+COS  c  (cos  a— cos  b)  =m\  c  (sin  b  cos  A— sin  a 

cos  B)  ; 

which  can  be  put  in  the  forms 

(1 — cose)  (cos  a -|- cos  J) = sin  c  (sin  b  cos  A -|- sin  a  cos  B), 
(1 +COS  c)  (cos  a— cos  J)  =  sin  c  (sin  b  cos  A — sin  a  cos  B) ; 

multiplying  these  equations  together,  substituting  for  (1  — cos') 
its  value  (sin'),  and  for  (cos')  its  value  (1  — sin'),  and  dividing  by 
Bill*  c,  we  have, 


SPHERICAL   TRIGONOMETRY. 


207 


COS*  a— cos'  5=sin''  5— sin'  h  sin"  A— sin'  a+sin*  a  sin'  B; 
then,  since  (1—  sin'a)— (1—  sin=  h)  =sin'  h  —  sin' «,  we  have, 

cos'  a  —  cos*  6=sin'  h  —  sin'  a,  and  we  get, 
sin"  h  sin'  Am  sin'  a  sin'  B  ; 

extracting  tlie  sqiiare  root,   sin  5  sin  A = sin  a  sin  B,  or  ' 
sin  A     sin  a 


sin  B     sin  b' 
sin  A 


derive 


And  from  first  and  third  formulae  (l)  we 
From  the  second  and  third  formu- 


sin  C     sin  c 

sin  C     sin  c 


\i?) 


lae  we  derive  —. — t^_  .    ,. 
sin  ±>     sin  0 

Or,  In  every  spherical  triangle^  the  sines  of  the  angles  are 
to  each  other  as  the  sines  of  their  opposite  sides. 

8.  Taking  the  third  formula  (l),  and  substituting  for  (cos  h)  its 
value  as  expressed  in  the  second,  and  for  (cos'  a)  its  value 
(1— sin'  a),  and  dividing  by  sin  a,  we  will  have, 

cos  c  sin  a=sin  c  cos  a  cos  B  +  sin  b  cos  C. 

But  (Spher.  Trig.  7)  we  have  sin  b~       ._  ^, — ;  substituting  for 


sin  C 


cose 


sin  b  its  value,  and  dividing  by  sin  c,  we  get,  -. sin  a=C08  a 


cosB4 


sin  B  cos  C 


sin  C 


_,      cos 

But     -r-  =COt, 

sin 


Hence  we  can  derive,  by  similar  processes, 

cot  a  sin  b=cot  A  sin  C  +  cos  b  cos  C, 
cot  a  sin  c=cot  A  sin  B+cos  c  cosB, 
cot  b  sin  a=cot  B  sin  C  +  cos  a  cosC, 
cot  b  sin  c=cot  B  sin  A+cos  c  cos  A, 
cot  c  sin  a=:cot  C  sin  B+cos  a  cos  B, 
cot  c  sin  b=cot  C  sin  A+cos  b  cos  A. 


}      (4) 


The  formulae  (1)  are  the  fundamental  analogies  of  Spherical 
Trigonometry,  from  which  all  the  others  are  derived,  which 
others  are  more  adapted  for  logarithmic  computations. 


208  THE   ELEMENTS   OF 


THE    SOLUTION    OF   RIGHT-ANGLED    SPHERICAL   TRIANGLES    BY 

LOGARITHMS, 

The  following  equations  give  the  unknown  parts  of  a  right- 
angled  spherical  triangle  when  C  is  the  right  angle  and  any 
two  other  parts  are  known.     There  are  six  cases. 

Let  C  be  the  right  angle,  and  c  be  the  hypothenuse. 

Case  1.  Given  a  and  b  to  find  c,  A,  and  B  ; 

,              .      tan  a           _,     tnn  b 
cos  e=cos  a  cos  o;  tan  A= r;  tan  l3= . 

■^  tan  o  tan  a 

Case  2.  Given  c  and  a  side  b  to  find  a,  A,  and  B ; 

cos  c  .      tan  b       .    „     sin  & 

cos  a= r;  cosA=: — — :   sin  13=^ . 

cos  b  tan  c  sin  c 

Case  3.  Given  aside  a  and  opposite  angle  A  to  find  others; 

.     ,     tan  a       .     „     sin  a       .     _     cos  A 

sm  0= — r  ;  sin  C=-. — -;  sin  B= -. 

tan  A  sm  A  cos  a 

Both  acute  or  both  not  acute. 

Case  4.  Given  a  side  a  and  adjacent  angle  B  to  find  others; 

tan  6=sin  a  tan  B ;  cot  c=:cot  a  cos  B  ;  cos  A  =  cos  a  sin  B. 

Case  5.    Given   the  hypothenuse  c  and  an  angle  A  to  find 
others ; 
sin  «  =  sine  sin  A ;  tan  5i=tan  c  cos  A;  cot  B  =  cose  tan  A. 

Case  6.  Given  the  oblique  angles  A  and  B  to  find  others ; 

cos  A  -     cos  B  .  _ 

cos  a=—. — ^rr;  cos  o=— — -  ;  cos  c  =  cot  A  cot  B. 
sin  B  sin  A 

Napier's  circtdar  parts  are  much  the  simplest  method  of  re- 
solving right-angled  spherical  triangles;  they  are  the  two 
sides  about  the  right  angle,  the  complements  of  the  oblique 
angles,  and  the  complement  of  the  hyjiothenuse.  Hence  there 
are  five  circular  parts;  the  right  angle  not  being  a  circular 
part,  is  supposed  not  to  separate  the  two  sides  adjacent  to  the 
right  angle ;  therefore  these  sides  are  regarded  adjacent  to 
each  other,  so  that  when  any  two  parts  are  given,  tlieir  corre- 
sponding circular  ])aits  are  also  known,  and  these  with  the  re- 
quired part  constitute  the  three  parts  under  consideration ; 
therefore  these  three  paias  will  lie  together,  or  one  of  tliem 


SPHERICAL   TRIGONOMETRY. 


209 


■will  be  separated  from  both  tlie  others.  Hence  one  part  is 
known  as  the  middle  part;  and  when  three  parts  are  undei-  con- 
sideration, the  parts  separated  by  the  middle  part  are  caUcd 
t\\G  adjacent  parts ;  and  the  parts  separated  liuni  tlie  middle 
parts  are  called  the  opposite  parts.     IS'ow,  assume  any  part  in 


B 


the  diagram  for  the  middle  part,  and  using  the  formulae  (1) 
•when  the  other  parts  are  the  ojiposite  parts,  and  we  get,  27ie 
sine  of  the  middle  part  is  equal  to  the  product  of  the  cosines  of 
the  opposite  parts.  Then,  assume  again  any  part  for  the  mid- 
dle, and  use  the  formulae  (2)  when  the  other  parts  are  the  ad- 
jacent parts,  and  we  get,  The  sine  of  the  middle  pa.  t  is  equal 
to  the  pyroduct  of  the  tangents  of  the  adjacent  parts.  Hence 
"We  derive  the  five  following  equations  : 

sin  a=tan  h  tan  (90"— B)=cos  (90° — A)  cos  (90°  — c;, 
sin  6=tan  a  tan  (90°— A)=:cos  (90°— B)  cos  (90°— c), 
sin  (90°— A)=tan  b  tan  (90°— c)=cos  (90°— c)  cos  c/, 
sin  (90° — c)=tan  (90°— A)  tan  (90°— B)==cos  a  cos  6,. 
sin  (90°— B)— -tana  tan  (90°— c)=cos  6  cos  (90° — A). 

The  a7za7o^/es  of  Xapier  are  derived  from  the  formulce  (l)  by 
eliminating  the  cosines  of  any  of  the  sides,  reducing  and  chang- 
ing to  linear  sines  and  cosines  (Plane  Trig.  p.  193),  and  we 
bave, 

cos  I  {a-\-h)  :  cos  |  (a — b) : :  cot  |  C  :  tan  |  (A  +  B), 
sin  \  (a+b)  :  sin  -^  (a—b)  : :  cot  -^  C  :  tan  ^  (A— B), 
cos  -^  (a-f  c)  :  cos  |  {a—c) : :  cot  ^  B  :  tan  ^  (A+C), 
sin  ^  {a+c)  :  sin  -^  (a—c)  : :  cot  |  B  :  tan  ^  (A — C), 
cos  ^  {b-hc)  :  cos  -^  (b—c) : :  cot  ^  A  :  tan  ^  (B  +  C), 
sin  ^  {b+c)  :  sin  -^  {b—c)  : :  cot  |  A  :  tan  |  (B — C), 
14 


210  THI.   ELEMENTS   OF 

The  same  proportions  applied  to  tlic  triangle  polar  to  ABC, 
with  accents  omitted,  we  have, 

cos  I  (A  +  B)  :  cos  i  (A— B)  : :  tan  |  c  :  tan  i  {a+b), 
sin  i  (A+B)  :  sin  i  (A— B)  : :  tan  -J  c  :  tan  i  (a-b), 
cos  i  (A+C)  :  cos  i  (A— C)  : :  cot  ^  &  :  tan  ^  (a+c), 
sin  i  (A+C)  :  sin  |  (A— C)  : :  cot  i  6  :  tan  ^  (a— c), 
cos  i  (B  +  C)  :  cos  ^  (B— C)  : :  cot  i  a  :  tan  i  (6+c), 
sin  i  (B  +  C)  :  sin  i  (B— C)  : :  cot  i  a  :  tan  |  (*— c). 

The  same  ambiguity  that  there  is  between  plane  triangles 
(I  3,  Fourth  Case)  exists  also  between  spherical  triangles, 
which  may  be  avoided  by  remembering  that  every  angle  and 
gide  of  a  spherical  triangle  are  each  less  than  two  right  angles, 
and  that  the  greater  angle  is  opposite  to  the  greater  side,  and 
the  least  angle  is  opposite  to  the  least  side;  and  conversely. 

Quadrantal  spherical  triangles  are  such  which  have  one  side 
equal  to  ninety  degrees ;  hence  they  can  very  easily  be  solved 
by  formulae  for  right-angled  spherical  triangles. 

SOLUTION   OF   OBLIQUE-ANGLED   SPHERICAL  TRIANGLES  BT 

LOGARITHMS. 

Case  1.  Given  the  three  sides  to  find  the  angles. 

Ain  {s—fi)  sin  (5— 5)  sin  (»— c) 
Find  s=i  (a+6+c)  ;let  M=|/ ■ ^-'^  ^ 5 

then, tan iA  =  ^-^^^^3:^;  tan  i  B=^^^^^-;   tan  i  C= 

]\I 

ein  (.« — c)' 

Case  2.  Given  two  sides  a  and  b  and  the  included  angle  C, 

to  find  others. 

Tan^(A+B)=^^^^i^4NotiC;  tan^A-B^'-i^^-^ 
■^^"^^^^^^     cosi(a+6)        '     '  Biui(u+6) 

cot  \  C. 

But  A=i  (A+B)+i  (A-B) ;  B=|  (A+B)-i  (A-B)  ; 

sin  c=sin  a^!^=sin  b  ^^,  or  find  an  angle  cot  9=tan  o 
sin  A  sin  15 

cos  a  sin  (J+(j)) 

cos  C ;  cos  0= -. • 

'  sin  (J) 


BPOEKICAL   TKIGONOMETKY. 


211 


Case  3.  Given  the  sides  a  and  b  and  an  angle  opposite  to 
one  of  them,  to  find  others. 

Find  cot  (p=tan  b  cos  A,  and  tanj^^cos  b  tan  A;  then,  sin 

,    ,    .      cos  a  sin  (p     .    ^       .     .    sin  b     .     ,    ,     .  . 

{c+(p)=i — ;  sm  JL>  =  sm  A  -; —  :  sin  (c +>-)=:  cot  a  tan  0 

^        '  cos  6     '  sin  a  ^      ^' 

fiinx- 

Case  4.  Given  the  angles  A  and  B  and  the  inchiaed  side,  to 

find  others. 

Find  tan  (p=cos  c  tan  A,  and  tan  p^=cos  c  tan  B ; 

,  tan  c  sin  (p  ,      tan  c  sin  y 

then,  tan  a—  .     .,,  , — -;  tan  6  =  -.    ,  .    ,    t; 

p_cos  A  cos  (B  +  (p)  _  cosB  cos  (A+y) 
cos  9  cos  -/^ 

Case  5.  Given  A  and  B  and  a  side  opposite  one  of  them,  to 
find  others. 

Find  tan  (p=:tan  a  cos  B ;  cot  ;^=:C08  a  tan  B; 


ein  irzsin  a 


sin  B 
sin  A  ■ 


sin  (<3— (p)=cot  A  tan  B  sin  9, 


•    //-I       \      cos  A  sin  Y 

sm  (C — y)  = — ^. 

^       ^'  cosB 

Case  6.  Given  the  three  angles  A,  B,  C,  to  find  the  sides. 
Take       • 

S=|(A+B  +  C);andN=,4/ — -        ""'*'''  ^ 

'     cos  10 — J 


cos  (S— A)  cos  (S— B)  cos  (tS— C) ; 

then,  tan  \  a—1^  cos  (S  —  A) ; 
tan|  &=N' cos  (S  — B); 
tan  \  c=N  cos  (S  —  C). 


*  THE   SURFACE   OP   A    SPHERICAL  TRIANGLE. 

Let  ABC  be  a  spherical  triangle,  AC=: 
DF,  BC=FE,  ABC=DEF. 

S=surface  of  ACB,  5=surface  of  hemis- 
phere—BHDC—AGEC—DCE=0  R'— 
(lune  AHD  —  S)  — (lune  BGE—  S)— (lune 
CDFE-S) 


=eR'(,-j 


80 


180"""  W  +  ^^- 


212  BPHKRICAL   TRIGONOMETKT. 

180*^,  or  equivalent  to  the  sum  of  the  three  angles  above  180°; 
hence,  its  spherical  excess  is  sometimes  taken  as  the  measure 
of  the  ti'iangle. 

Or  in  terms  of  its  sides,  formula  given  by  L'Huiller, 

tan  ^  E=  V  [tan  ^  s  tan  ^  {s  —  a)  tan  ^  (s  —  b)  tan  ^  {s  —  c)]. 


EXERCISES  IN 

ELEMENTARY    GEOMETRY, 

AND  m 

PLANE  AND  SPIIEKICAL  TKIGONOMETRY.* 


DEFI^nXIOXS. 


1.  Lines^  avgUs^  and  spaces  are  said  to  be  given  in  magni- 
tude, when  they  are  either  exhibited,  or  when  the  method  of 
finding  them  is  known. 

2.  Points,  lines,  and  spaces  are  said  to  be  given  in  position, 
■which  have  always  the  same  situation,  and  which  are  actually 
exhibited  or  can  be  found. 

3.  A  circle  is  said  to  be  given  in  magnitude  when  its  radius 
is  oiven  ;  and  \w position,  when  its  center  is  given. 

Mao-nitudes,  instead  of  being  said  to  be  given  in  magnitudey 
or  given  in  position,  are  often  said  simply  to  be  given,  when  no 
ambiguity  arises  from  the  omission. 

*  For  these  Exercises  I  am  indebted  to  Thomson's  Euclid  (Belfast),  they 
being  judiciously  selected  by  that  eminent  writer,  and  their  presentation 
here  is  a  valuable  acquisition  to  an  American  school  text-book.  I 
would  gladly  acknowledge  my  obligations  for  many  propositions  in 
this  volume,  but  they  being  culled  for  more  than  two  thousand  years 
from  the  best  writers  on  Geometry,  and  being  so  much  modified  by  each 
succeeding  age,  that  it  is  impossible  at  this  day  to  attribute  tliem  to  their 
rightful  authors,  and  many  of  them  being  more  or  less  contained  in 
every  work  on  the  subject,  they  have  become  public  property.  What  I 
have  introduced  myself  will  be  well  recognized  by  every  student  of 
Geometry,  and  my  only  apolugy  is,  the  desire  to  advance  the  cause  of 
Truth. 


214  EXEBCISK8. 

4.  A  ratio  is  said  to  be  giveti  when  it  is  the  same  as  that  of 
two  cciven  maGrnitudes. 

5.  A  rectilineal  figure  is  said  to  be  given  in  species,  when  its 
several  anarles  and  the  ratios  of  its  sides  are  iriven. 

6.  When  a  series  of  unequal  magnitudes,  unlimited  in  num- 
ber, agree  in  certain  relations,  the  greatest  of  them  is  called  a 
ma/x'mum ^  the  least,  a  minimum. 

Thus,  of  chords  in  a  given  circle,  the  diameter  is  the  maxi- 
mum ;  and  of  straiiiht  lines  drawn  to  a  given  straight  line, 
from  a  given  point  without  it,  the  minimum  line  is  the  perpen- 
dicular. 

7.  A  line  which  is  such  that  any  point  whatever  in  it  fulfills 
certain  conditions,  is  called  the  locus  of  that  point. 

8cho.  1.  Several  instances  of  loci  have  already  occurred  in 
the  preceding  books. 

1.  Tlius,  it  was  stated  in  the  fifth  corollary  to  the  fifteenth 
proposition  of  the  first  book,  that  all  triangles  on  tlie  same 
base,  aiid  between  the  same  pai'allels,  are  equivalent  in  area; 
and  hence,  if  only  the  base  and  area  of  a  triangle  be  given,  its 
vertex  may  be  at  any  point  in  a  straight  line  parallel  to  the 
base,  and  at  a  distance  from  it  which  may  be  determined  by 
ap])lying  (11.  5,  scho.)  to  half  the  base  a  parallelogram  equiva- 
lent to  the  given  area ;  and  therefore  the  parallel  is  the  locus 
of  the  vertex.  Here  the  conditions  fulfilled  are,  that  straight 
lines  drawn  from  any  point  in  the  parallel  to  the  extremities  of 
the  given  line,  form  with  it  a  triangle  having  a  given  area. 

2.  It  was  stated  in  the  first  corollary  to  the  eighteenth  proposi- 
tion of  the  third  book,  that  all  anHes  in  the  same  segment  of  a 
cinOe  are  equal ;  and  hence,  if  only  the  base  and  vertical  angle 
of  a  triangle  be  given,  the  vertex  may  be  at  any  point  of  the 
arc  of  a  segment  described  on  the  base,  in  the  manner  pointed 
out  in  the  nineteenth  pi'oposition  of  the  third  book;  that  arc, 
therefore,  is  the  locus  of  the  vei-tex. 

3.  It  will  be  seen  in  the  sixth  proposition  of  these  Exercises  that 
straight  lines  drawn  from  any  point  whatever  in  the  circumfer- 
ence of  the  circle  ABGC  to  the  points  E,  F,  have  the  same 
ratio — that  of  EA  to  AF.  Hence,  therefore,  if  the  base  of  a 
triangle,  and  the  ratio  of  the  sides  be  given,  the  locus  of  the 
■vertex  is  the  circumference  of  the  circle  described  in  the  man- 


EXEUCISES.  215 

ner  pointed  out  in  the  corollary  to  this  proposition ;  unless 
the  ratio  be  that  of  equality,  in  which  case  the  locus  is  evi- 
dently a  perpendicular  bisecting  the  straight  line  joining  the 
points. 

4,  It  follows  likewise,  from  the  fifth  corollary  to  the  twcnty- 
fomth  proposition  of  the  first  book,  that  when  the  base  of  a  tri- 
angle and  the  difference  of  the  squares  of  the  sides  are  given, 
if  the  point  D  be  found  (II.  12,  scho.)  in  the  base  BC,  or  its 
continuation,  such  that  the  difference  of  the  squares  of  BD,  CD 
is  equivalent  to  the  difference  of  the  squares  of  the  sides;  and 
if  through  D  a  perpendicular  be  drawn  to  ]3C,  straight  lines 
drawn  from  any  point  of  the  perpendicular  to  B,  C  will  have 
the  difference  of  their  squares  equivalent  to  the  given  difference; 
and  hence  the  perpendicular  is  the  locus  of  the  vertex,  when  the 
base  and  the  difference  of  the  squares  of  the  sides  are  given. 

5.  It  will  appear  in  a  similar  manner  from  the  corollary  to  the 
twelfth  proposition  of  the  second  book,  that  if  BC  the  base  of  a 
triangle,  and  the  sum  of  the  squai'es  of  the  other  sides  AB,  AG 
be  given,  the  locus  of  the  vertex  is  the  circumference  of  a  circle 
described  from  D,  the  middle  point  of  the  base  as  center,  and 
with  the  radius  T>A.  To  find  DA,  take  the  diagonal  of  the 
square  of  BD  as  one  leg  of  a  right-angled  triangle,  and  for  its 
bypothenuse  take  the  side  of  a  square  equivalent  to  the  given 
sum  of  the  squares  of  AB,  AC  ;  then  the  diagonal  of  the  square 
described  on  half  the  remaininoj  le<x  of  that  rigcht-aniiled  trian- 
gle  will  be  the  radius  of  DA.  The  proof  of  this  is  easy,  de- 
pending on  the  third  and  fourth  coi'ollaries  to  the  twenty-fourth 
jDroposition  of  the  first  book,  and  on  the  corollai-y  to  the  twelfth 
proposition  of  the  second  book. 

Scho.  2.  In  discovering  loci,  as  well  as  in  other  investigations 
in  geometry,  the  stuflent  is  assisted  by  what  is  termed  geomet- 
rical analysis  ;  of  the  nature  of  which  it  may  be  proper  here 
to  give  some  explanation. 

Take  this  proposition  :  If  a  chord  of  a  given  circle  have  one 
extremity  given  in  positior),  and  if  a  segment  terminated  at 
that  extremity  he  taken  on  the  ch  rd,  produced  if  necessary^ 
such  that  the  rectangle  under  the  segment  and  chord  may  be 
equivalent  to  a  given  space  ;  the  locus  of  the  point  of  section  is 
a  straight  I i?ie  given  inp)osition. 


216  EXERCISES. 

Let  AB  be  the  diameter  of  the  circle  and  AC  a  chord  of  the 
circle. 

If,  in  the  proposition,  instead  of  being  informed  that  the  locns 
is  a  strai!j;ht  line,  we  were  required  to  find  what  tlie  locus  is, 
we  might  proceed  in  the  following  manner:  Lft  D  be  any  point 
in  the  required  line,  so  that  the  rectangle  ACAD  is  equivalent 
to  the  given  space  ;  and  having  drawn  the  diameter  AB,  find 
E,  so  that  the  rectangle  AB.AE  may  be  equal  to  ACAD,  and 
therefore  E  a  point  in  the  required  line  ;  and  join  DE,  BC 
Then  (V.  10,  cor.)  AB  :  AC  :  :  AD  :  AE.  Hence  (V.  6)  tho 
triangles  DAE,  BAC,  having  the  angle  A  common,  are  equi- 
angular; and  thereiore  AED  is  equal  to  ACB,  which  is  a  right 
angle.  The  point  D  is  therefore  in  a  perpendicular  passing 
through  E;  and  in  the  same  manner  it  would  be  t^hown,  that 
any  other  point  in  the  required  line  is  in  the  perpendicular ; 
that  is,  the  perpendicular  is  its  locus. 

The  investigation  just  given  is  called  the  analysis  of  the 
projjosition,  while  the  solutions  hitherto  given  are  called  the 
synthesis  or  compositio7i.  In  analysis  we  commence  by  sup- 
posing that  to  be  effected  which  is  to  be  done,  or  that  to  be 
true  which  is  to  be  proved  ;  and,  by  a  regular  succession  of  con- 
sequences founded  on  that  supposition,  and  on  one  another, 
we  arrive  at  something  which  is  known  to  be  true,  or  which 
we  know  the  means  of  effecting.  Thus,  in  the  second  corollary 
to  the  seventeenth  proposition  of  the  sixth  book,  the  conclusion 
obtained  for  the  area  of  the  circle  is  shown  by  the  third  corol- 
lary to  be  consistent  with  the  proportion  established  by  Ar- 
chimedes between  the  cone,  sphere,  and  cylinder,  and  also 
consistent  with  the  geometrical  truth  in  regard  to  the  sur- 
faces of  the  sphere  and  cylinder.  Plence,  analysis  takes  into 
consideration  this  consistence,  and  confirms  the  second  corol- 
lary from  its  agreement  with  established  truths  of  geometr3^ 

Again:  in  the  corollary  to  the  twenty-fourth  proposition  of 
the  fifth  book,  since  circles  are  in  propoition  to  the  squares  of 
tlieir  radii,  the  quadrant  ACB  is  equivalent  to  the  semicircle 
ADC,  we  have  (I.  ax.  3)  the  triangle  ABC  equivalent  to  th^ 
crescent  ADC.  Now,  when  we  api)ly  the  conclusion  derived 
by  the  second  corollary  to  the  seventeenth  proposition  of  the 
sixth  book  to  the  above,  we  find  a  perfect  agreement ;  taking 


EXERCISES.  ^17 

the  circle  as  three  times  square  of  radius,  we  have  quadrant 
ACB  equivalent  to  J  AB';  hence  (I.  ax.  1)  the  semiciicle  ADO 
is  equivalent  to  |  AW.  But  (VI.  17,  cor,  2)  the  segment  AC 
of  the  quadrant  ACB  is  equivalent  toi  AB";  therefore  (I.  ax.  3) 
we  have  the  triangle  ABC  and  the  crescent  ADC  each  equiva- 
lent to  ^  AB',  thus  showing  the  agreement  between  the  second 
corollary  to  the  seventeenth  proposition  of  the  sixth  book,  and 
the  established  truth  relating  to  the  crescent  or  lune. 

Also,  we  have  (VI.  17,  cor.  2)  the  hemisphere  generated  by 
the  quadrant  BNP  equivalent  to  the  solid  generated  by  the 
trapezium  BSNP,  and  we  have  the  solid  generated  by  the 
figure  BTNP  common  ;  thei-efore  (I.  ax.  3)  the  solid  generated 
by  the  segment  BT  is  equivalent  to  the  solid  generated  by  the 
figure  TSN.  Now,  the  solid  generated  by  the  segment  BT  is 
a  part  of  the  hemisphere;  hence  its  contents  are  computed  by 
the  same  radius  as  the  hemisphere  ;  the  solid  generated  by  the 
figure  TSN  is  a  part  of  the  solid  generated  by  the  trapezium 
BSNP ;  hence  its  contents  are  also  computed  by  the  same 
radius  as  the  hemisphere.  Therefore  we  obtain  by  analysis 
the  ti-uth,  that  vchen  equivalent  solids  are  generated  by  equiva- 
lent surfaces,  the  generating  surfaces  are  up  n  equal  radii,  a 
truth  corresponding  to  the  truth  established  by  the  second 
coroUaiy  to  the  seventeenth  proposition  of  the  sixth  book,  tJiat 
equivalent  surfaces  upon  the  same  radius  loill  generate  equiva- 
lent solids.  The  synthesis  then  commences  with  the  conclusion 
of  the  analysis,  and  retraces  its  sevei-al  steps,  making  that  pre- 
cede which  before  followed,  till  we  arrive  at  the  required  con- 
clusion. Therefore  the  demonstrations  given  in  the  second 
corollary  to  the  seventeenth  proposition,  book  sixth,  obtain  the 
conclusion  from  which  the  analyses  precede.  From  this  it  ap- 
pears that  analysis  is  the  instrument  of  investigation ;  while 
svnthesis  affords  the  means  of  communicatino-  what  is  already 
known ;  and  hence,  in  the  Elements  of  Euclid,  the  synthetic 
method  is  followed  throughout.  What  is  now  said  will  receive 
further  illustration  from  the  solution  of  the  following  easy 
problem. 

Given  the  perimeter  and  angles  of  a  triangle,  to  construct  it. 

Analysis. — Suppose  ABC  to  be  the  required  triangle,  and 
produce  BC  both  ways,  making  BD  equal  to  BA,  and  CE  to 


218 


EXERCISES. 


CA ;  then  DE  is  given,  for  it  is  equal  to  the  sum  of  the  throo 
sides  AB,  BC,  CA ;  that  is,  it  is  equal  to  the  given  perimeter. 
j^  Join  AD,  AE.     Then  (I.  1,  cor. 

1)  the  angles  D  and  DAB  are 
equal,  and  therefore  each  of 
them  is  half  of  ABC,  because  (T. 
20)  ABC  is  equal  to  both.  The 
angle  D  therefore  is  given  ;  and 
in  the  same  manner  it  may  be  shown  that  E  is  given,  being 
half  of  ACB.  Hence  the  triangle  ADE  is  given,  because  the 
base  DE,  and  the  angles  D,  E  are  given  ;  and  ADE  being 
given,  ABC  is  also  given,  the  angle  DAB  being  equal  to  D, 
and  EAC  equal  to  E. 

Composition. — Make  DE  equal  to  the  given  perimeter, 
the  angle  D  equal  to  the  half  of  one  of  the  given  angles,  and  E 
equal  to  the  half  of  another;  draw  AB,  AC,  maldng  the  angle 
DAB  equal  to  D,  and  EAC  to  E;  ABC  is  the  triangle  re- 
quired. 

For  (T.  1,  cor.  2)  AB  is  equal  to  BD,  and  AC  to  CE.  To 
these  add  BC,  and  the  three,  AB,  BC,  CA,  are  equal  to  DE, 
that  is,  to  the  given  perimeter.  Also  (I.  20)  the  angle  ABC  is 
equal  to  D  and  DAB,  and  is  therefore  double  of  D,  since  D 
and  DAB  are  equal. .  But  D  is  equal  to  the  half  of  one  of  the 
given  angles;  therefore  ABC  is  equal  to  that  angle;  and,  in 
the  same  manner,  ACB  may  be  proved  to  be  equal  to  another 
of  the  given  angles.  ABC  therefore  is  the  triangle  required, 
since  it  has  its  perimeter  equal  to  the  given  perimeter,  and  its 
angles  equal  to  the  given  angles. 

It  is  impossible  to  give  rules  for  effecting  analyses  that  will 
answer  in  all  cases.  It  may  be  stated,  however,  in  a  general 
way,  that  when  sums  or  differences  are  concerned,  the  corre- 
sponding sums  or  differences  should  be  exhibited  in  the  as- 
sumed figure ;  that  in  many  cases  remarkable  points  should  be 
joined  ;  or  that  tlirough  them  lines  may  be  drawn  perpendicu- 
lar or  parallel  to  remarkable  lines,  or  making  given  angles  with 
them  ;  and  that  circles  may  be  described  with  certain  radii, 
and  from  certain  points  as  centers;  or  touching  certain  lines, 
or  passing  tlirough  certain  points.  Some  instances  of  analysis 
will  be  given  in  subsequent  propositions ;  and  the  student  will 


EXERCISES.  219 

find  it  useful  to  make  analyses  of  many  other  propositions,  such 
as  several  in  the  Exercises. 

8.  A  jyorism  is  a  proposition  affirming  the  possibility  of  find- 
ing such  conditions  as  will  render  a  certain  problem  indeter- 
minate, or  capable  of  innumerable  solutions. 

Scho.  3.  Porisms  may  be  regarded  as  liaving  their  origin  in 
the  solution  of  j^roblenis,  which,  in  particular  cases,  on  account 
of  pccidiar  relations  in  their  data,  admit  of  innumerable  solu- 
tions ;  and  the  proposition  announcing  the  property  or  relation 
which  renders  the  problem  indeterminate,  is  called  a  porism. 
This  will  be  illustrated  by  the  solution  of  the  following  easy 
problem. 

Through  a  given  point  A,  let  it  be  required  to  draw  a  straight 
line  bisecting  a  given  parallelogram  BCDE. 

Suppose  AFG  to  be  the  required  line,  and  let  it  cut  the  sides 
BE,  CD  in  F,  G,  and  the  diagonal  CE  in  H.  Then  from  tho 
equivalent  figures  EBC,  FBCG  take  FBCII,  and  the  remaining 
triangles  EHF,  ClIG  are  equal.  Now,  since  (I.  16  and  11) 
these  triangles  are  equiangular,  it  is  evident  that  they  can  be 
equal  in  area  only  when  their  sides  are  equal ;  wherefore  II  is 
the  middle  point  of  the  diagonal.  The  construction,  therefore, 
is  effected  by  bisecting  the  diagonal  EC  in  H,  and  drawiu"' 
AFHG.  For  the  triangles  CHG,  EIIF  are 
equiangular,  and  since  CH,  HE  are  equal,  the 
triangles  are  equal.  To  each  of  them  add  the 
figure  FBCH  ;  then  the  figure  FBCG  is  equiv- 
alent to  the  triangle  EBC,  that  is,  to  half  the 
parallelogram  BD. 

Now,  since  the  diagonal  CE  is  given  in 
magnitude  and  position,  its  middle  point  II 
is  given  in  position,  and  therefore  H  is  always  a  point  in  the 
required  line,  wherever  A  is  taken.  Hence,  so  long  as  A  and 
II  are  different  points,  the  straight  line  AHG  (I.  post  1)  is  de- 
termined. This,  however,  is  no  longer  so,  if  the  given  point  A 
be  the  intersection  of  the  diagonals,  that  is,  the  point  H,  as  in 
that  case  only  one  point  of  the  required  line  is  known,  and  the 
problem  becomes  indetermhiate,  any  straight  line  whatever, 
through  H,  equally  answering  the  conditions  of  the  problem; 
and  we  are  thus  led  by  the  solution  of  the  problem  to  the  con- 


220  EXERCISES. 

elusion,  that  in  a  parallelogram  a  point  may  he  found,  such 
that  a)iy  straight  line  wliatever  drawn  through  it,  bisects  the 
parallelogra7n  /  and  this  is  a  porism. 

The  seventy-sixtii  pi'oposition  of  the  Exercises,  when  con- 
sidered in  a  particular  manner,  affords  another  instance  of  a 
porism  ;  as  it  appears  that  if  a  circle  and  a  point  D  or  E  be 
given,  another  point  E  or  D  may  be  found,  such  that  any  circle 
whatever,  desciibed  through  D  and  E,  will  bisect  the  circum- 
ference of  the  given  circle;  and  this  may  be  regarded  as  the 
indeterminate  case  of  the  problem,  in  which  it  is  required, 
through  two  given  points,  to  describe  a  circle  bisecting  tlie  cir- 
cumference of  another  given  circle, — a  problem  which  is  always 
determinate,  except  when  the  points  are  situated  in  the  man- 
ner supposed  in  the  proposition. 

9.  Isopjeritnetrical  figures  are  such  as  have  their  perimeters, 
or  bounding  lines,  equal. 

10,  The  general  problem  of  the  tangencies,  as  understood  by 
the  ancients,  is  as  follows:  Of  three  points,  three  straight  lines, 
and  three  circles  of  given  radii,  any  three  being  given  in  posi- 
tion ;  it  is  required  to  describe  a  circle  passing  through  the 
points,  and  touching  the  straight  lines  and  circles.  This  gen- 
eral problem  comprehends  ten  subordinate  ones,  the  data  of 
which  are  as  follows:  (l.)  three  points;  (2.)  two  points  and  a 
straight  line ;  (3.)  two  points  and  a  circle ;  (4.)  a  point  and  two 
straight  lines  ;  (5.)  a  point,  a  straight  line,  and  a  circle  ;  (G.)  a 
point  and  two  circles;  (7.)  three  straight  lines;  (8.)  two 
straight  lines  and  a  circle  ;  (9.)  a  straight  line  and  two  circles; 
and  (10.)  three  circles.  The  first  and  seventh  of  these  are  the 
second  and  fifth  corollaries  of  the  twenty-fifth  proposition  of 
the  third  book. 

If  a  circle  be  continually  diminished,  it  may  be  regarded  as 
becoming  ultimately  a  point.  By  being  continually  enlarged, 
on  the  contrary,  it  may  have  its  curvature  so  much  diminished 
that  any  portion  of  its  circumference  may  be  made  to  differ  in 
as  small  a  degree  as  we  please  from  a  straight  line.  Viewing 
the  subject  in  this  light,  we  may  regard  the  first  nine  of  the 
problems  now  mentioned,  as  comprehended  in  the  tenth.  Thus, 
we  shall  have  the  first,  by  supposing  the  circles  to  become  in- 
finitely small ;  the  seventh,  by  supposing  them  infinitely  great; 


EXEHCISES. 


221 


the  fifth,  by  taking  one  of  them  infinitely  small,  one  infinitely 
great,  and  one  as  a  circle  of  finite  magnitude;  and  so  on  with 
regard  to  the  othei'S.  These  views  of  tlie  subject  tend  to  illus- 
trate it ;  but  they  do  not  assist  in  the  solution  of  the  problems. 
Hcho.  4.  In  tiie  fifth  problem,  the  straight  line  may  fall  with- 
out the  circle,  may  cut  it,  or  may  touch  it ;  the  point  may  be 
without  the  circle,  withfn  it,  or  in  its  circumference  ;  or  it  may 
be  in  the  given  straight  line,  or  on  either  side  of  it ;  and  it  will 
be  an  interesting  exercise  for  the  student,  in  this  and  many 
other  problems,  to  consider  the  variations  arising  in  the  solu- 
tion from  such  changes  in  the  relations  of  the  data,  and  to  de- 
termine what  relations  make  the  solution  possible,  and  what 
render  it  impossible.  It  may  also  be  remarked,  that  in  many 
problems  there  will  be  slight  variations  in  the  proofs  of  differ- 
ent solutions  of  the  same  problem,  even  when  there  is  no 
chi"ige  in  the  method  of  solution;  such  as  in  the  present  in- 
stance, when  the  required  circle  is  touched  externally,  and 
Avhen  internally.  Thus,  while  in  one  case  angles  may  coincide, 
in  another  the  corresponding  ones  may  be  vertically  ojiposite; 
and  the  reference  luay  sometimes  be  to  the  coJiverse  of  the  first 
corollary  and  sometimes  to  the  converse  of  the  second  corollary 
of  the  eighteenth  proposition,  book  third.  It  is,  in  general,  un- 
necessary to  point  out  these  variations,  as,  though  they  merit 
the  attention  of  the  student,  they  occasion  no  difficulty. 

PROPOSITIONS. 

Prop.  I. — Tiieor. — If  an  angle  of  a  triangle,  he  bisected  by  a 
straight  line^  which  likewise  cuts  the  base,,  the  rectangle  con- 
tained by  the  sides  of  the  triangle  is  equivalent  to  the  rectangle 
contained  by  the  segments  of  the  base,,  together  with  the  square 
of  the  straight  line  bisecting  the  angle. 

Let  ABC  be  a  triangle,  and  let  the  angle  BAC  be  bisected 
by  AD;  the  rectangle  BA.AC  is  equal  to  the  rectangle  BD.DC, 
together  with  the  square  of  AD. 

Describe  the  circle  (III.  25,  cor.)  ACB  about  the  triangle ; 
produce  AD  to  meet  the  circumference  in  E,  and  join  EC. 
Then  (hyp.)  the  angles  BAD,  CAE  are  equal ;  as  are  also  (IIL 
18,  cor.  1)  the  angle  B  and  E,  for  they  are  in  the  same  segment; 


222  EXERCISES. 

thej-efore  (V.  3,  cor.)  in  the  ti-iangles  ABD,  AEC,  as  BA  :  AD 
::  EA  :  AC;  and  consequently  (V.  10,  cor.)  the  rectangle  BA. 
AC  is  equivalent  to  EA.AD,  that  is  (II.  3),  ED.DA,  together 
with  the  square  of  AD.     But  (III.  20)  the  rectangle  ED.DA  ia 


equivalent  to  the  rectangle  BD.DC ;  therefore  the  rectangle 
BA.AC  is  equivalent  to  BD.DC,  together  with  the  square  of 
AD ;  wherefore,  if  an  angle,  etc. 

Scho.  From  this  proposition,  in  connection  -with  the  fourth 
proposition  of  fifth  book,  we  have  the  means  of  computing  AD, 
when  the  sides  are  given  in  numbers.  For, by  the  fourth,  BA: 
AC  ::  BD  :  DC;  and,  by  composition,  BA+AC  :  AC  : : 
BC  :  DC.  This  analogy  gives  DC,  and  BD  is  then  found  by 
taking  DC  from  BC.  But  by  this  proposition  BA.  AC = BD.DC 
+  AD-;  therefore  from  BA.AC  take  BD.DC,  and  the  square 
root  of  the  remainder  will  be  AD, 

In  a  similar  manner,  from  the  fourth  proposition  of  the  fifth 
book,  and  the  second  proposition  of  these  Exercises,  the  line 
bisecting  the  exterior  vertical  angle  may  be  computed;  and 
from  the  third  proposition  of  these  Exercises,  in  connection 
with  the  eleventh  or  twelfth  of  the  second  book,  the  diameter 
of  the  circumscribed  circle  may  be  computed,  when  the  sides 
of  the  triangle  are  given  in  numbers. 

Prop.  II. — Theor. — If  an  exterior  angle  of  a  triangle  be 
bisected  by  a  straight  line,  which  cuts  tlie  base  produced  ; 
the  rectangle  contained  by  the  sides  of  the  triangle,  and  the 
square  of  the  bisecting  line  are  together  equivalent  to  the  rect- 
angle contained  by  the  segments  of  the  base  intercepted  between 
its  extremities  and  the  bisecting  line. 

Let,  in  the  foregoing  diagram,  the  exterior  angle  ACF  of  the 
triangle  BAC  be  bisected  by  HC ;  the  rectangle  BC.  AC  and 


EXERCISES.  223 

the   square  of  HC  are   together  equivalent  to  the  rectangle 
BH.AII. 

Describe  the  circle  (III.  25,  cor.)  ABEC  about  the  triangle 
BAG;  produce  HC  (I.  post.  2)  to  E,  and  join  EA.  Then, 
since  (l>yp.)  the  angles  FCII  and  HCA  are  equal,  their  supple- 
ments FCE  and  ACE  (T.  def.  20  and  ax.  3)  are  also  equal ;  and 
(III.  18,  cor.  1)  B  and  E  are  equal.  Therefore  (V.  3,  cor.)  in 
the  triangles  BCH  and  EAC,  BC  :  CH  :  :  EC  :  AC,  and  con- 
sequently (V.  10,  cor.)  the  rectangle  BC.  AC  is  equivalent  to 
EC.CII.  To  each  add  square  of  CII;  therefore  BC.AC+CIP 
are  equivalent  to  EC.CH+Cff;  or  (II.  3)  BC.AC  +  CIP  are 
equivalent  to  EH.CH  ;  or  (III.  21,  cor.)  BC.AC  +  CIP  are 
equivalent  to  BH.  AH.     Therefore,  if  an  exterior  angle,  etc. 

Prop.  III. — Theor. — If  from  an  angle  of  a  triangle  a  per- 
pendicular be  drawn  to  the  basej  the  rectangle  contained  by 
the  sides  of  the  triangle  is  equivalejit  to  the  rectangle  contained 
by  the  perpendicular  and  iAe  diameter  of  the  circle  described 
about  the  triangle. 

Also,  in  the  foregoing  diagram,  let  ABC  be  a  triangle,  AL 
the  perpendicular  from  the  angle  A  to  BC ;  and  AE  a  diameter 
of  the  circumscribed  circle  ABEC;  the  rectangle  BA.AC  is 
equivalent  to  the  rectangle  AL.AE. 

Join  EC.  Then  the  right  angle  BLA  is  equal  (III.  11)  to  the 
angle  EC  A  in  a  semicircle,  and  (HI.  18,  cor.  1)  the  angle  B  to 
the  angle  E  in  the  same  segment ;  therefore  (V.  3,  cor.)  as 
BA  :  AL  :  :  EA  :  AC;  and  consequently  (V.  10,  cor.)  the 
rectangle  BA.AC  is  equivalent  to  the  rectangle  EA.AL.  If, 
therefore,  from  an  angle  of  a  triangle,  etc. 

Prop.  IV. — Theor. — The  rectam^le  contained  by  the  diago- 
nals of  a  quadrilateral  inscribed  in  a  circle,  is  equivalent  to 
both  the  rectangles  contained  by  its  opposite  sides. 

Let  ABCD  be  a  quadrilateral  inscribed  in  a  circle,  and  join 
AC,  BD ;  the  rectangle  AC.BD  is  equivalent  to  the  two  rect- 
angles AB.CD  and  AD.BC. 

Make  the  angle  ABE  equal  to  DBC,  and  take  each  of  them 
from  the  whole  angle  ABC ;  then  the  remaining  angles  CBE, 
ABD  are  equal;  and  (HI.  18,  cor.  l)  the  angles  ECB,  ADB 
are  equal.     Therefore  (V.  3,  cor.)  in  the  triangles  ABD,  EBC, 


224 


EXERCISES. 


as  BC  :  CE 
BD.CE 


Again 


BD  :  DA;  wlience  (V.  10,  cor.)  BC.DA= 
in  the  triangles  BAE,  BDC,  because  (const.) 
the  angles  ABE,  DBC  are  equal,  as  also 
(III.  18,  cor.  1)  BAE,  BDC;  therefore 
(V.  3,  cor.)  as  BA  :  AE  :  :  BD  :  DC; 
^\  hence  (V.  10,  cor.)  BA.DC==BD.AE. 
'  Add  these  equivalent  rectangles  to  the 
equivalents  BC.DA  and  BD.CE;  then 
BA.DC  +  BC.DA  =  BD.CE  +  BD.AE,  or 
(II.  1)  BA.DC  +  BC.DA=BD.AC.  There- 
fore, the  rectangle,  etc. 
Cor.  1.  If  the  sides  AD,  DC,  and  consequentlj^  (TIT.  16,  cor. 
1)  the  arcs  AD,  DC,  and  the  angles  ABD,  CBD  be  equal,  the 
rectangle  BD.AC  is  equivalent  to  AB.AD  together  with  BC. AD, 
or  (II.  1)  to  the  rectangle  under  AD,  and  the  sum  of  ABand  BC. 
Hence  (V.  10,  cor.)  AB-f-BC  :  BD  :  :  AC  :  AD  or  DC. 

Cor.  2.  If  AC,  AD,  CD  be  all  equal,  the  last  analogy  be- 
comes AB+BC  :  BD  :  :  AD  :  AD ;  whence  AB  +  BC=BD. 
Hence  in  an  equilateral  triangle  inscribed  in  a  circle,  a  straight 
line  drawn  from  the  vertex  to  a  point  in  the  arc  cut  oiF  by  the 
base  is  equal  to  the  sum  of  the  chords  drawn  from  that  point 
to  the  extremities  of  the  base. 


Prop.  V. — Theor. — The  diagonals  of  a  qtiadrilateral  in- 
scribed  in  a  circle,  are  proportional  to  the  sums  of  the  rectan- 
gles contained  by  the  sides  meet'nr/  at  their  extremities. 

Let  ABCD  be  a  quadi-ilateral  inscribed  in  a  circle,  and  AC, 
BD  its  diagonals;  AC  :  BD  : :  BA.AD-h 
BC.CD  :  AB.BC+AD.DC. 

If  AC,  BD  cut  one  another  perpendicu- 
larly in  L,  then  (Ex.  2)  AK  being  the  diame- 
ter of  the  circle,  AL.  AK  =  BA.AD,  and  CL. 
AK=:BC.CD;  whence,  by  addition  (I.  ax. 
2),  AL.AK  +  CL.AK,  or  (II.  1)  AC.AKz= 
B.\.AD+BC.CD.  In  a  similar  manner, 
it  would  be  proved  that  BD.AK  =  AB.BC 
Hence  ACIAK  :  BD.AK,  or  (V.  1)  AC  :  BD  :  • 
BA.AD  +  BC.CD  :  AB.BC  +  AD.DC. 

But  if  AC  be  not  perpendicular  to  BD,  draw  AEF  perpen- 


+  AD.DC. 


EXERCISES. 


225 


dicnlar  and  CF  parallel  to  BD,  and  DGH  perpendicular  and 
BH  parallel  to  AC.     Then,  because  EF  is  equal  to  the  perpen- 
dicular drawn  from  C  to  BI),  and  GH  equal  to  the  one  drawn 
from  B  to  AC ;  it  would  be  prov- 
ed as  before,  that  AP\AIv  =  BA. 
AD  +  BC.CD,     and     I)n.AK  = 
AB.BC  +  AD.DC.      Hence,   AF. 
AK  :    DH.AK,  or    (V.    l)   AF  : 
DII   ::  BA.AD-^BC.CD  :   AB. 
BC  +  AD.DC.     But  the  triani^les 
AFC,  DIIB  are  equiangular,  liav- 
in<x  the  rio;ht  anojes  Fand  H,  and 
the  angles  ACF,  DBH,  each  equal 
(I.  16)  to  ALD;  therefore  (V.  3) 

AF  :  AC  :  :  DH  :  DB,  and  alternately  AF  :  T3IT  :  :  AC  : 
DB.  Hence  (IV.  7)  the  foregoing  analogy  becomes  AC  :  BD 
:  :  BA.AD  +  BC.CD  :  AB.BC-hAD.DC.  Wherefore,  the 
diagonals,  etc. 

Scho.  From  this  proposition  and  the  last,  when  the  sides  of 
a  quadrilateral  inscribed  in  a  circle  are  given,  we  can  find  the 
ratio  of  the  diagonals  and  their  rectangle,  and  thence  (V,  15) 
the  diagonals  themselves.  Also,  if  the  sides  be  given  in  num- 
bers, we  can  compute  the  diagonals.  Thus,  let  the  sides  taken 
in  succession  round  the  figure  be  50,  78,  104,  and  120.  Then, 
the  ratio  of  the  diagonals  will  be  that  of  50  x  784-104  x  120  to 
60x120-1-78x104;  that  is,  16880  to  14112,  or  65  to  56,  by- 
dividing  by  252.  Again,  the  rectangle  of  the  diagonals  is  50  x 
104-1-78x120,  or  14560.  But  similar  rectilineal  figures  are  as 
the  squares  of  the  corresponding  sides,  and  consequently  the 
sides  are  as  the  square  roots  of  the  areas ;  therefore,  taking  65 
and  56  as  the  sides  of  a  rectangle,  we  have  its  area  equal  to 
3640;  and  ^  3640  is  to  4/14560,  or  -^Z  3640  is  to  |/ (4x3640), 
that  is,  1   :  2  :  :  65  :  130  :  :  56 


112  ;  so  that  130  and  112 


are  the  diagonals. 


Prop.  VI. — Theor. — If  in  a  straight  line  drawn  through 

the  center  of  a  circle^  and  on  the  same  side  of  (he  center,  (wo 

povits  be  taken  so  that  the  radius  is  a  mean  proportional  her- 

tween  their  distances  from,  the  cei^ter  /  two  straight  lines  drawn 

15 


226  EXEECISES. 

from,  those  points  to  any  point  whatever  in  the  circumference^ 
are  proportional  to  the  segments  into  which  the  circumference 
divides  the  straight  line  intercepted  leticeen  the  same  points. 

Let  ABC  l)e  a  circle,  and  CAE  a  straight  line  drawn  throngh 
its  center  D ;  if  ED  :  DA  :  :  DA  :  DF,  and  if  BE,  BF  be 
drawn  from  any  point  B  of  the  circumference ;  EB  :  BF  :  : 
EA  :  AF. 

Join  AB,  BD,     Then,  since  DB  is  equal  to  DA,  we  have 

(hyp.)  ED  :  DB  ::  DB  :  DF. 
The  two  triangles  EDB,  BDF, 
therefore,  have  their  sides  about 
the  common  angle  D  proportion- 
al ;  wherefore  (V.  6)  the  angle 
E  is  equal  to  FBD.  Now  the 
angle  liAD  is  equal  (I.  20)  to  the 
two  angles  E  and  EBA,  and  also 
(I.  1,  cor.)  to  ABD ;  wherefore  E  and  EBA  are  (I.  ax.  1)  equal 
to  ABD.  From  these  take  the  equal  angles  E  and  FBD,  and 
(I.  ax.  3)  the  remaining  angles  EBA,  ABF  are  equal ;  and 
therefore  (V.  4)  in  the  triangle  EBF,  EB  :  BF  :  :  EA  :  AF. 
If,  therefore,  in  a  straight  line,  etc. 

Cor.  Join  BC,  and  produce  EB  to  G.  Then,  since  ABC  is 
(III.  11)  a  right  angle,  it  is  equal  to  the  two  EBA,  CBG. 
From  these  equals  take  the  equal  angles  ABF,  EBA,  and  the 
remainders  FBC,  CBG  are  equal ;  and  therefore  (V.  4,  2d  case) 
EB  :  BF  :  :  EC  :  CF.  But  it  has  been  proved  that  EB  : 
BF  :  :  EA  :  AF  ;  therefore  (IV.  1)  EA  :  AF  :  :  EC  :  CF. 
Hence,  if  the  segments  EA,  AF  be  given,  the  point  C  may  be 
determined  by  the  method  shown  in  the  third  corollary  to  the 
sixteenth  proposition  of  the  first  book;  and  the  circle  ABC 
may  then  be  described,  its  diameter  AC  being  determined. 

Scho.  The  circle  may  also  be  determined  in  the  following 
manner:  Since  (hyp.)  ED  :  DA  :  :  DA  :  DF,  by  division, 
EA  :  DA  :  :  AF  :  DF ;  whence,  alternately  and  by  division, 
EA— AF  :  AF  :  :  AF  :  DF.  Hence  DF  is  a  third  proper- 
tional  to  the  difference  of  E A,  AF,  and  to  AF,  the  less ;  and 
thus  the  center  D  is  determined.  From  the  last  analogy  also 
we  obtained  (IV.  11)  EA— AF  :  EA  :  :  AF  :  AD;  an  anal- 
ogy which  serves  the  same  purpose,  since  it  shows  that  the  ra- 


EXERCISES.  227 

dius  of  the  circle  is  a  fourth  proportional  to  the  difference  of 
the  segments  EA,  AF,  and  to  those  segments  themselves. 

Prob.  VII, — TnEOR.  —  The  perpendiculars  drawn  from  the 
three  angles  of  any  triangle  to  the  ojjposite  sides,  intersect  one 
another  in  the  same  point. 

If  the  triangle  be  right-angled,  it  is  plain  that  all  the  perpen- 
diculars pass  through  the  right  angle.  But  if  it  be  not  right- 
angled,  let  ABC  be  the  triangle,  and  about  it  describe  a  circhi; 
then,  B  and  C  being  acute  angles,  draw  ADE  perpendicular  to 
BC,  cutting  BC  in  D,  and  the  circumference  in  E;  and  make 
DF  equal  to  DE ;  join  BF  and  produce  it,  if  necessary,  to  cut 
AC,  or  AC  produced  in  G ;  BG  is  perpeixlicular  to  AC.  Join 
BE ;  and  because  FD  is  equal  to  DE,  the  angles  at  D  right 
angles,  and  DB  common  to  the  two  triangles  FDB,  EDB,  the 
angle  FBD  is  equal  (I.  3)  to  EBD ;  but 
(III.  18,  cor.  1)  CAD,  EBD  are  also  equal, 
because  they  are  in  the  same  segment  ; 
therefore  CAD  is  equal  to  FBD  or  GBC, 
But  the  angle  ACB  is  common  to  the  tv  o 
triangles  ACD,  BCG;  and  therefore  (I, 
20,  cor.  5)  the  remaining  angles  ADC, 
BGC  are  equal ;  but  (const.)  ADC  is  a 
right  angle;  therefore  also  BGC  is  a  right  angle,  and  BG  is 
perpendicular  to  AC,  In  the  same  manner  it  would  be  shown 
that  a  straight  line  CII,  drawn  through  C  and  F,  is  perpendicu- 
lar to  AB.  The  three  perpendiculars  therefore  all  pass  through 
F ;  wherefore,  the  perpendiculars,  etc, 

Scho.  This  limitation  prevents  the  necessity  of  a  different 
ease,  which  would  arise  if  the  perpendicular  AD  fell  without 
the  triangle.  If  the  angle  A  be  obtuse,  the  point  F  lies  with- 
out the  circle,  and  BF,  not  produced,  cutsAC  produced.  The 
proof,  however,  is  the  same,  and  it  is  very  easy  and  obvious. 
Another  easy  and  elegant  proof,  of  which  the  following  is  an 
outline,  is  given  in  Garnier's  "Reciproques,"  etc.,  Theor.  III., 
page  78 :  Draw  BG  and  CH  perpendicular  to  AC  and  AB ; 
join  GH,  and  about  the  quadrilaterals  AHFG  and  BHGC  de- 
8cribe  circles,  which  can  be  done,  as  is  easily  shown  ;  draw 
also  AFD.     Then  the  angles  BAD,  BCH  are  equal,  each  of 


223  EXERCISKS. 

them  being  equal  (TIT,  18,  cor.  1)  to  HGF ;  and  the  angle  ABO 
"being  coniinoii,  ADB  is  equal  (I.  20)  to  BHC,  and  is  therefore 
a  right  angle. 

Prop.  VTTT. — Theor. — From  AB,  the  greater  side  of  the  tri- 
angle ABC,  cut  off  AD  equal  to  AC,  and  join  DC;  draw  AE 
bisecting  the  vertical  angle  BAC,  andjoiji  DE ;  d  aw  also  AF 
perpendicular  to  BC,  and  DG  parallel  to  AE.  Then  (1.)  the 
angle  DEB  is  eqtdvalent  to  the  difference  of  the  angles  at  the 
base,  ACB,  ABC  ;  or  of  BAF,  CAF,  or  of  AEB,  AEC  ;  and 
DE  is  equal  to  EC  ;  (2.)  the  angles  BCD,  EAF  are  each  equiv- 
alent to  half  the  same  difference  ;  (3.)  ADC  or  ACD  is  equiva- 
lent to  half  the  sum  of  the  angles  at  the  base,  or  to  the  comple- 
ment of  half  the  vertical  angle ;  (4.)  BG  is  equivalent  to  the 
difference  of  the  segments  BE,  EC,  made  by  the  line  bisecting 
the  vertical  angle.  ^ 

1.  In  the  triangles  AED,  AEC,  AD,  AC  are  equal,  AE  com- 
mon, and  the  contained  angles  equal ;  therefore  (I.  3)  DE  is 
equal  to  EC,  the  angle  ADE  to  ACE,  and  AED  to  AEC.    But 

(I.  20)  because  BD  is  produced,  the 
angle  ADE  is  equivalent  to  B  and 
BED;  therefore  BED  is  the  differ- 
ence of  B  and  ADE,  or  of  B  and 
ACB.  Also  BED  is  the  difference 
of  AEB,  AED,  or  of  AEB,  AEC. 
Again  :  ABF,  BAF  are  equivalent 
to  ACF,  CAF,  each  pair  being  (L 
20,  cor.  3)  equivalent  to  a  right  angle.  Take  away  ABF; 
then,  because  the  difference  of  ACF,  ABF  is  BED,  there  re- 
mains BAF  equivalent  to  BED,  CAF ;  that  is,  BED  is  the 
difference  of  BAI\  CAF. 

2.  The  difference  BED  is  equivalent  (T.  20)  to  the  two  angles 
ECD,  EDC,  which  (I.  1,  cor.)  are  equal ;  therefore  ECD  is  half 
of  BED.  Again  :  in  the  triangles  AlID,  AIIC,  because  AD, 
AC  are  equal,  AIT  common,  and  the  contained  angles  equal, 
DII  is  equal  (I.  3)  to  IIC,  and  the  angles  at  H  are  equal,  and 
are  therefore  right  angles.  Then,  in  the  triangles  AEF,  CEII, 
the  angles  AFE,  CUE  are  equal,  and  AEC  common ;  therefore 


EXERCISES.  229 

(T.  20,  cor.  5)  the  article  EAF  is  equal  to  ECU,  which  has  been 
proved  to  be  equivalent  to  the  half  of  BED, 

3,  Since  the  anole  BAG  is  common  to  the  triangjles  ABC, 
ADC,  the  angles  ADC,  ACD  are  (I.  20)  equal  to  ABC,  ACB ; 
and  therefore,  since  ADC,  ACD  are  equal,  each  of  them  is 
equivalent  to  half  the  sum  of  ABC,  ACB  ;  also  either  of  them, 
ADC,  is  the  complement  ofDAIi,  half  the  vertical  angle,  since 
AHD  is  a  right  angle. 

4.  Because  DH,  HC  are  equal,  and  HE,  DG  parallel,  GE  is 
equal  (V.  2)  to  EC;  and  therefore  BG,  the  difference  of  BE, 
GE,  is  also  the  difference  of  BE,  EC. 

Sclw.  1.  It  is  easy  to  prove  without  proportion,  that  if  AB  (in 
the  figure  for  the  above  proposition)  be  bisected  in  D,  the 
straight  line  DE  parallel  to  BC  bisects  AC,  and  that  the  trian- 
gle ADE  is  a  fourth  of  ABC.  For  (I.  15,  cor.  5)  the  triangles 
BDC,  BEC  are  equal.  But  (I.  15,  cor.  5)  BDC  is  half  of  ABC  ; 
and  therefore  BEC  is  half  of  ABC,  and  is  equal  to  BEA. 
Hence  the  bases  AE,  EC  are  equal ;  for  if  they  were  not,  the 
triangles  ABE,  CBE  (I.  15,  cor.  6)  would  be  unequal.  Again  : 
because  AD,  DB  are  equal,  the  triangle  ADE  is  (I.  15,  cor.  5) 
half  of  ABE,  and  therefore  a  fourth  of  ABC. 

Conversely,  if  DE  bisect  AB,  AC,  it  is  parallel  to  BC.  For 
(T.  15,  cor.)  the  triangles  BDC,  BEC  are  each  half  of  ABC; 
and  these  being  therefore  equal,  DE  is  parallel  (I.  15,  cor.  5)  to 
BC. 

Hence,  it  is  plain  (I.  14)  that  the  straight  lines  joining  DE, 
DF,  EF  divide  the  triangle  ABC  into  four  equal  triangles, 
similar  to  the  whole  and  to  one  another;  and  that  each  of 
these  lines  is  equal  to  half  the  side  to  which  it  is  parallel. 

Scho.  2.  Instead  of  cutting  off  AD  equal  to  AC,  AC  may  be 
produced  through  C,  and  by  cutting  off,  on  AC  thus  produced,  a 
part  terminated  at  A,  and  equal  to  AB,  and  by  making  a  con- 
struction similar  to  that  of  the  foregoing  proposition,  it  will  be 
easy  to  establish  the  same  properties  as  those  above  demon- 
strated, or  ones  exactly  analogous. 

Prop.  IX. — Prob. —  Giveii  the  base  of  a  triangle,  the  differ- 
ence of  the  S'des,  and  the  difference  of  the  angles  at  the  base; 
to  construct  it. 


230  EXERCISES. 

Make  BC  equal  to  the  given  base,  and  CBD  equal  to  half 
the  difference  of  the  angles  at  the  base ;  from  C  as  center,  at  a 
distance  equal  to  the  difference  of  the  sides,  describe  an  arc 
cutting  BD  in  D  ;  join  CD  and  produce  it ;  make  the  angle 
DBA  equal  to  BDA  ;  ABC  is  the  required  triangle. 

For  (I.  1,  cor.)  AD  is  equal  to  AB,  and  the  difference  of  AC, 
AD,  or  of  AC,  AB,  is  CD ;  and  (Ex.  8) 
since  AD  is  equal  to  AB,  CBD  is  equal 
to  half  the  difference  of  the  angles  at  the 
base.  The  triangle  ABC,  therefore,  is 
the  one  required,  as  it  has  its  base  equal 
to  the  given  base,  the  difference  of  its 
sides  equal  to  the  given  difference,  and 
the  difference  of  the  angles  at  the  base  equal  to  the  given  dif- 
ference of  those  angles. 

Method  of  C\miputation.  In  the  triangle  BCD  there  are 
given  BC,  CD,  and  the  angle  CBD ;  whence  the  angle  C  can 
be  computed  ;  and  the  sum  of  this  and  twice  CBD  is  the  angle 
ABC.  Then,  in  the  whole  triangle  ABC,  the  angles  and  the 
Bide  BC  are  giveti ;  whence  the  other  sides  may  be  computed  ; 
or,  one  of  them  being  computed,  the  other  will  be  found  by 
means  of  the  given  difference  CD. 

Prop.  X.— Prob. —  Given  the  segments  into  which  the  base 
of  a  triangle  is  divided  by  the  line  bisecting  the  vertiecd  anglQ 
and  the  difference  of  the  sides  ;  to  construct  the  triangle. 

Construct  the  triangle  CED,  having  the  sides  CE,  ED  equal 
to  the  given  segments,  and  CD  equal  to  the  given  dillerence  of 
the  sides;  produce  CE,  and  make  EB  equal  to  ED;  bisect  the 
angle  BED  by  EA,  meeting  CD  produced  in  A,  and  join  AB ; 
ABC  is  the  required,  triangle. 

A  For,  in  the  triangles  AEB,  AED,  BE 

is  equal  to  ED,  EA  common,  and  the 
angle  BEA  equal  to  DEA  ;  therefore  (I, 
3)  BA  is  equal  to  DA,  and  the  angle 
EAB  to  EAD.  Hence,  ABC  is  the  re- 
quired trianale;  for  CD,  the  difference 
of  its  sides,  is  equal  to  the  given  differ- 
ence, and   BE,    EC,    the   segments   into    which    the   base  is 


FXERCISE8. 


231 


divided  by  the  line  bisecting  the  vertical  angle,  are  equal  to 
the  given  segments. 

Meth  d  of  Computation.  The  sides  of  the  triangle  CDE 
are  given,  and  therefore  its  angles  may  be  computed;  one  of 
which  and  the  sni)plement  of  the  other  are  tlie  angles  C  and 
B.  Then,  in  the  triangle  ABC,  the  angles  and  BC  are  given, 
to  compute  the  remaining  sides. 

OTHERWISE, 

Since  (V.  4)  CE  :  EB  :  :  CA  :  AB,  we  have,  by  division, 
CE— EB  :  EB  :  :  CA— AIJ  :  AB  ;  which,  therefore,  becomea 
known,  since  the  first  three  terms  of  the  analogy  are  given; 
and  thence  AC  will  be  found  by  adding  to  AB  the  given  dit- 
ference  of  the  sides. 

Prop.  XT. — I  V.ob. —  Given  the  base  of  a  triangle^  the  vertical 
angle,  and  the  difference  of  the  sides ;  to  construct  the  tri- 
angle. 

Let  MNO  be  the  given  vertical  angle;  produce  ON"  to  P, 
and  bisect  the  angle  MNP  by  NQ.  Then,  make  BD  equal  to 
the  difference  of  the  sides,  and  the 
angle  ADC  equal  to  QNP;  from  B 
as  center,  with  the  given  base  as  ra- 
dius, describe  an  arc  cutting  DC  in 
C ;  and  make  the  angle  DCA  equal 
to  ADC  ;  ABC  is  the  required  tri- 
angle. For  (const.)  the  angles  ACD, 
ADC  are  equal  to  MNP,  and  there- 
fore (I.  20  and  9)  the  angle  A  is 
equal  to  MN).  But  (I.  1,  cor.)  AD 
is  equal  to  AC,  and  therefore  BD  is 
the  difference  of  the  sides  AB,  AC  ; 
and  the  base  BC  is  equal  to  the 
given  base  ;  wherefore  ABC  is  the  triangle  required. 

Method  of  Computation.  In  the  triangle  CBD,  the  sides 
BC,  BD,  and  the  angle  BDC,  the  supplement  of  ADC  or  MNQ 
are  given ;  whence  the  other  angles  can  be  computed.  The 
rest  of  the  operation  will  proceed  as  in  the  ninth  proposition  of 
these  Exercises. 


Q 


N 


232  EXERCISES. 

Prop.  XII. — Prob. —  Given  one  of  the  angles  at  the  base  of 
a  triangle,  the  difference  of  the  sides,  and  the  difference  of  the 
segments  into  which  the  base  is  divided  by  the  line  bisecting  the 
vertical  angle;  to  construct  the  triangle. 

Construct  the  triangle  DBG,  liaving  the  angle  B  equal  to 
the  given  angle,  BD  equal  to  the  difference  of  the  sides,  and 

BG  equal  to  the  difference  of  the  segments ; 
draw  DC  perpendicular  to  DG,  and  meet- 
^'^  '^  ing  BG  produced  in  C;  produce  BD,  and 

make  the  angle  DCA  equal  to  CDA  ; 
B  G  E  c  ABC  is  the  triangle  required.  For  it  has 
B  equal  to  the  given  angle,  and  the  differ- 
ence of  its  sides  BD  equal  to  the  given  difference;  and  if  AHE 
be  drawn  bisecting  the  angle  BAC,  it  bisects  (I.  3)  CD,  and  is 
perpendicular  to  it ;  it  is  therefore  parallel  to  DG,  one  side  of 
the  triangle  CDG;  and,  bisecting  CD  in  II,  it  also  (V,  2) 
bisects  GC  in  E.  Hence  BG,  the  difference  of  BE,  GE  is  also 
the  difference  of  BE,  EC,  the  segments  into  which  the  base  is 
divided  by  the  line  bisecting  the  vertical  angle. 

Method  of  Computation.  In  the  triangle  DBG,  BD,  BG, 
and  the  angle  B  are  given  ;  whence  (Trig.  3)  we  find  half  the 
difference  of  the  angles  BGD,  BDG,  which  is  equal  to  half  the 
angle  C.  Then  (by  the  same  proposition)  we  have  in  the  tri- 
angle ABC,  tan  I  (C-B)  :  tan  ^  (C+B)  :  :  c-b  or  BD  :  c+ 
b  ;  whence  the  sides  c  and  b  become  known,  and  thence  BC 
by  the  first  case. 

For  (I.  20)  BEA=r|  A+C,  and  consequently  BEA— |  A  =  C. 
But  (I.  16)  BGD  =  BEA,  and  BDG^BAE^^A;  and  there- 
fore BGD— BDG =C. 

Prop.  XIII. — Prob. —  Given  the  base  of  a  triangle,  the  verti- 
cal angle,  and  the  sutn  of  the  sides  ;  to  construct  it. 

Make  BD  equal  to  the  sum  of  the  sides,  and  the  angle  D 
equal  to  half  the  vertical  angle ;  from  B  as  center,  with  the 
base  as  radius,  describe  an  arc  meeting  DC  in  C ;  and  make 
the  angle  DCA  equal  to  D  ;  ABC  is  the  required  triangle. 

For  (I.  1,  cor.)  AD  is  equal  to  AC,  and  therefore  BA,  AC  are 
equal  to  BD,  the  given  sum.  Also  (I.  20)  the  exterior  angle 
BAC  is  equal  to  the  two  D  and  ACD,  or  to  the  double  of  D, 


EXERCISES. 


233 


because  D  and  ACD  are  equal ;  therefore,  since  D  is  half  the 

given  vertical  angle,  BAG  is  equal 

to  that  ansjle.     The  triancjle  ABC, 

therefore,  has  its  base  equal  to  the 

given  base,  its  vertical  angle  equal 

to  the  given  one,  and  the  sum  of  its 

sides  equal  to  the  given  sum ;  it  is 

therefore  the  triangle  required. 

Scho.  Should  the  circle  neither  cut  nor  touch  DC,  the  prob- 
lem would  be  impossible  with  the  proposed  data.  If  the  circle 
meet  DC  in  two  points,  there  will  be  two  triangles,  each  of 
which  will  answer  the  conditions  of  the  problem.  These  tii- 
angles,  however,  will  differ  only  in  position,  as  they  will  be  on 
equal  bases,  and  will  have  their  remaining  sides  equal,  each  to 
each.  This  problem  might  also  be  solved  by  describing  (III. 
19)  on  the  given  base  15C  a  segment  of  a  circle  containing  an 
angle  equal  to  half  the  vertical  angle ;  by  inscribing  a  chord 
BD  equal  to  the  sura  of  the  sides;  by  joining  DC;  and  then 
proceeding  as  before.  The  construction  given  above  is  prefer- 
able. / 

Method  of  Computation.  In  the  triangle  BDC,  the  angle 
D,  and  the  sides  BC,  BD  are  given  ;  whence  the  remaining 
angles  can  be  compute  1 ;  and  then,  in  the  triangle  ABC,  the 
angles  and  the  side  BC  are  given,  to  compute  the  other  sides. 


Prop.  XIV. — Prob. —  Given  the  vertical  angle  of  a  triangle^ 
and  the  segments  into  which  the  line  iisectiiig  it  divides  the 
base  ;  to  construct  it. 

In  the  straight  line  BC,  take  BH  and  CII  equal  to  the  seg- 
ments of  the  base ;  on  BC  describe  (III. 
19)  the  segment  BAC  containing  an  angle 
equal  to  the  vertical  angle,  and  complete 
the  circle  ;  bisect  the  arc  BEC  in  E  ;  draw 
EHA,  and  join  BA,  CA ;  ABC  is  the  re- 
quired triangle.  For  (III.  16,  cor.  1)  the 
angles  BAII,  CAH  are  equal,  because  the 
arcs  BE,  EC  are  equal ;  and  therefore  the 
triangle  ABC  manifestly  answers  the  conditions  of  the  ques- 
tion. 


234  EXEECISF.8. 

Scho.  The  construction  might  also  be  effected  by  describing 
on  BH  and  CH  segments  each  containing  an  angle  equal  to  half 
the  vertical  angle,  and  joining  their  point  of  intersection  A 
■with  B  and  C.  Another  solution  may  be  obtained  by  the 
principle  (V.  4)  that  BA  •  AC  :  :  BII  :  HC.  For  if  a  trian- 
gle be  constructed  having  its  vertical  angle  equal  to  the  given 
one,  and  the  sides  containing  it  equal  to  the  given  segments, 
or  having  the  same  ratio,  that  triangle  will  be  similar  to  the 
required  one  ;  and  therefore  on  CB  construct  a  triangle  equi- 
angular to  the  one  so  obtained. 

Method  of  Computation.  Join  BE,  and  draw  ED  perpen- 
dicular to  BC.  Then  BD  or  DC  is  half  the  sum  of  the  sear- 
nients  BII,  HC,  and  DH  half  their  difference ;  and  BD  is  to 
DM,  or  twice  BD  to  twice  DH,  as  tan  DEB  to  tan  DEH. 
Now  it  is  easy  to  show  that  BED  is  half  the  sum  of  the  angles 
ABC,  ACB,  and  HED  half  their  difference  ;  and  therefore  these 
angles  become  known;  and  BC  being  given,  the  triangle  ABC 
is  then  resolved  by  the  method  for  the  Hrst  case. 

Cor.  Hence,  we  have  the  method  of  solving  the  problem  in 
wliich  the  base,  the  vertical  angle,  and  the  ratio  of  the  sides  of 
a  triangle  are  given^  to  construct  it.  For  (V.  4)  the  sides  be- 
ing proportional  to  the  segments  BH,  HC,  it  is  only  necessary 
to  divide  the  given  base  into  segments  proportional  to  the 
sides  and  then  to  proceed  as  above.  * 

Prop.  XV. — Pbob. —  Given  the  base^  the  perpendicular^  and 
the  vertical  angle  of  a  triangle  /  to  construct  it. 

Make  BC  equal  to  the  given  base,  and  (HI.  19)  on  it  de- 
scribe a  segment  capable  of  containing 
an  angle  equal  to  the  vertical  angle ; 
draw  AK  parallel  to  BC,  at  a  distance 
from  it  equal  to  the  given  perpendicular 
and  meeting  the  arc  in  A ;  join  AB, 
AC;  A13C  is  evidently  the  tiiangle  re- 
quired. 

Method  of  Compiitat'on.     Draw  the 

perpendicular    AD,   and    parallel   to  it 

draw  LGH,  through  the  center  G;  join  BG,  AG,  AH.     Now, 

iiuce  AH  evidently  bisects  the  anglu  BAC,  the  angle  HAD  or 


EXERCISES. 


235 


n  is  (Ex.  8)  equal  to  half  the  difference  of  the  ansrles  ABC, 
ACB,  and  therefore  (III.  1 0)  AGK  is  the  wliole  difference  of 
those  angles.  Then,  in  the  riglit-angled  triangle  BP'G,  the 
angles  and  BF  being  known,  FG  can  be  computed ;  from 
which  and  from  AD  or  KF,  KG  becomes  known.  Now,  to  the 
radius  BG  or  AG,  FG  is  the  cosine  of  BGF,  or  BAG,  and  KG 
the  cosine  of  AKG;  and  therefore  FG  :  KG  :  :  cos  liAC  : 
cos  AGK.  Hence  the  angles  ABC,  ACB  become  known,  and 
thence  the  remaining  sides. 

iSc/io.  Should  the  parallel  AK  not  meet  the  circle,  the  solu- 
tion  would  be  impossible,  as  no  triangle  could  be  constructed 
having  its  base,  perpendicular,  and  vevlic:il  angle  of  the  given 
magnitudes.  If  the  parallel  cut  the  cinrle,  there  will  be  two 
triangles,  either  of  which  will  answer  the  conditi(m  of  the  ques- 
tion. They  will  differ,  however,  only  in  position,  as  their  sides 
will  be  equal,  each  to  each.  If  the  parallel  touch  the  circle, 
there  will  be  only  one  triangle ;  and  it  will  be  isosceles. 


Pkop.  XVI. — Prob. —  Given  the  base  of  a  triangle,  the  ver- 
tical angle,  and  the  radius  of  the  inscribed  circle  ;  to  construct 
the  triangle. 

Let  GHK  be  the  given  angle  ;  produce  GH  to  L,  and  bisect 
LHK  by  HM;  on  the  given  base  BC  desciibe  (III.  19)  the  seg- 
ment BDC,  containing  an  angle  equal  to  GHM  ;  draw  a  straight 
line  parallel  to  BC,  at  a  distance  equal  to  the  given  ra<lius,  and 
meeting  the  arc  of  the  segment  in  D;  join  I)B,  DC;  and  make 
the  angles  DBA,  DCA  equal  to  DBC,  DCB,  each  to  each ; 
ABC  is  the  required  triangle.  L  ii  g 

Produce  BD  to  E,  and  drawDF  perpen- 
dicular to  BC.  Then,  since  (const.)  the 
angle  BDC  is  equal  to  GHM,  the  two  jj 
DBC,  DCB  are  equal  (I.  20  and  9)  to 
LHM,  and  therefore  (const.)  ABD,  ACD 
are  equal  to  KHM.  But  (1.  20)  BDC  is 
equal  to  BEC,  ECD,  or  to  BAC,  ABD, 
ACD,  because  (I.  20)  BEC  is  equal  to 
BAC,  ABD.  Therefore  BAC,  ABD,  ACD 
are  equal  to  GHM;  from  the  former  take 
ABD,  ACD,  and  from   the  latter  KHM,   which   is  equal  to 


236 


KXERCISE8. 


them,  and  the  remainders  BAG,  GHK  are  equal.  It  is  plain, 
also  (I.  14),  tliat  perpendiculars  drawn  IVoni  D  to  AB  and  AC 
would  be  each  equal  to  DF ;  and  therefore  a  circle  described 
from  D  as  center,  with  DF  as  radius,  would  be  inscribed  in  the 
trian<ijle  ABC  ;  and  BC  being  the  given  base,  and  A  being 
equal  to  the  given  vertical  angle,  ABC  is  the  required  tri- 
angle. 

The  method  of  computation  is  easily  derived  from  that  of  the 
preceding  proposition. 

Scho.  Should  the  parallel  to  BC  not  meet  the  arc  of  the  seg- 
ment, the  solution  would  be  impossible,  as  there  would  be  no 
triangle  which  could  have  its  base,  its  vertical  angle,  and  its 
inscribed  circle  of  the  given  magnitudes.  If  the  parallel  be  a 
tangent  to  the  arc,  the  radius  of  the  inscribed  circle  would  be 
a  maximum.  Hence,  to  solve  the  problem  in  which  the  base 
and  the  vertical  angle  are  given,  to  construct  the  triangle,  so 
that  the  inscribed  circle  may  be  a  maximum,  describe  the  seg- 
ment as  belbre,  and  to  find  D  bisect  the  arc  of  the  segment. 
The  rest  of  the  construction  is  the  same  as  before ;  and  the  tri- 
angle will  evidently  be  isosceles. 


Prop.  XVII. — Prob. —  Given  the  three  lines  drawn  from  the 
vertex  of  a  triangle,  one  of  them  'perpend  cnlorto  the  base,  one 
bisecting  the  base,  and  one  bisecting  the  vertical  angle  y  to  con- 
struct the  triitngle. 

Take  any  straight  line  BC  and  draw  DA  perpendicular  to  it, 
and  equal  to  the  given  perpendicular;  fi-om  A  as  center,  with 
radii  equal  to  the  lines  bisecting  the  vertical  angle  and  the 

base,  describe  arcs  cutting  BC  in  E  and 
F,  and  draw  AEH  and  AF  ;  through  F 
draw   GFII  perpendicular   to  BC,  and 
draw  AG  making  the  angle  IIAG  equal 
to  II,  and  cutting  HG  in  G;  from  G  as 
center,  with   GA   as  radius,  describe  a 
circle  cutting  BC  in  B  and  C;  join  AB, 
AC  ;  ABC  is  the  triangle  required. 
For  (HI.  2)  since  GFII  is  perpendicu- 
lar to  BC,  BC  is  bisected  in  F ;  and  (III.  17)  the  arcs  BII,  HO 
are  equal.     Therefore  (III.  16,  cor.  1)  the  angles  BAII,  CAH 


EXERCISES.  237 

are  equal.  Hence,  in  the  triangle  ABC,  the  perpendicular  AD, 
the  line  AE  bisecting  the  veitical  angle,  and  the  line  AF 
bisecting  the  base,  are  equal  to  the  given  lines.  Therefore 
ABC  is  the  triangle  required. 

Scho.  It"  the  three  given  lines  be  equal,  the  problem  is  inde- 
terminate ;  as  any  isosceles  triangle,  having  its  altitude  equal 
to  one  of  the  given  lines,  will  answer  the  conditions. 

3fethod  of  Computation.  Through  A  draw  a  parallel  to  BC, 
meeting  HG  ))rodnced  in  K,  Then,  in  the  right-angled  triangle 
ADE,  AE,  AD  being  given,  DAE,  or  H,  may  be  computed  ; 
the  double  of  which  is  AGK  ;  and  AK  or  FD  being  given,  AG, 
GK  can  be  found,  and  thence  GF.  Hence,  if  GB  were  drawn, 
it  and  GF  being  known,  BF,  and  the  angle  BGF,  or  BAC,  can 
be  computed.  The  rest  is  easy;  DAE,  half  the  difference  of  B 
and  C,  being  known. 

Analysis.  Let  ABC  be  the  required  triangle,  AD  the  per- 
pendicular, and  AE,  AF  the  lines  bisecting  the  vertical  angle 
and  the  base.  About  ABC  describe  (HI,  25,  cor.  2)  a  circle, 
and  join  its  center  G,  with  A  and  F,  and  produce  GF  to  meet 
the  circumference  in  H.  Then  (HI.  2)  GFH  is  perpendicular 
to  BC,  and  (IH.  \1)  the  arcs  BH,  HC  are  equal.  But  (HI.  16) 
the  equal  angles  BAE,  CAE  at  the  circumference  stand  on 
equal  arcs;  and  therefore  AE  being  produced,  will  also  pass 
through  H;  and  the  point  H,  and  the  angle  GHA  and  its  equal 
HAG  are  given.  Hence  also  the  center  G  and  the  circle  are 
given,  and  the  method  of  solution  is  plain. 

Prop.  XVIH. — Prob. —  Given  the  base  of  a  triangle^  the  ver- 
tical angle,  and  the  straight  line  bisecting  that  angle  y  to  con- 
struct the  triangle. 

On  the  given  base  BC  describe  (HI.  19)  the  segment  BAC 
capable  of  containing  an  angle  equal  to  the  given  vertical  an- 
gle, and  complete  the  circle ;  bisect  the  arc  BEC  in  E,  and  join 
EC;  perpendicular  to  this,  draw  CF  equal  to  half  the  line 
bisecting  the  vertical  angle,  and  from  F  as  center,  with  FC  as 
radius,  describe  the  circle  CGII,  cutting  the  straight  line  pass- 
in  through  E  and  F  in  G  and  H ;  make  ED  equal  to  EG,  and 
draw  EDA;  lastly,  join  AB,  AC,  and  ABC  is  the  required 
triangle. 


238  EXEKCI8E8. 

For  the  triangles  CEA,  CED  are  equiangular,  the  angle  CEA 
being  tomnion,  and  BCE,  CAE  being  each  equal  to  BAE. 
Therefuie  AE  :  EC  : :  EC  :  ED,  and  (V.  9,  cor.)  AE.ED= 


EC=.  But  (ITI.  16,  cor.  3,  and  21)  HE.EG  or  HE.ED=EC 
and  therefore  AE.ED^HE.ED;  whence  AE=:HE,  and  (I.  ax. 
3)  AD=GH  =  '2CF.  AD  is  therefore  equal  to  the  given  bisect- 
ing line,  and  it  bisects  the  angle  BAC.  Hence  ABC  is  the  re- 
quired triangle. 

Method  of  Computation.  Draw  EL  perpendicular  to  BC, 
and  join  CH.  Then  BCE  is  equal  to  BAE,  half  the  vertical 
angle  A ;  and  therefore,  to  the  radius  EC  ;  CL  is  the  cosine  of 
■§  A,  and  CF  is  the  tangent  of  CEF  to  the  same  radius  ;  where- 
fore, to  any  radius,  CL  :  CF,  or  BC  :  AD  :  :  cos  ^  A  :  tan 
CEF  or  cot  EFC ;  and  hence  the  angle  H,  being  half  of  EEC, 
is  known.  Also  ECU  is  the  complement  of  II,  because  ECF  is 
a  right  angle,  and  FCH  equal  to  H.  But  (Tgig.  2)  EC  :  EH  or 
EA  :  :  sin  II  :  sin  ECU,  or  cosH  ;  or  (Trig.  defs.  cor.  6)  EC  : 
EA  :  :  It  :  cot  H.  Also  in  the  triangle  ACE,  EC  :  EA  : : 
Bin  -^  A  :  sin  ACE ;  whence  (IV.  7)  R  :  cot  H  :  :  sin  ^  A  :  sin 
ACE ;  whence  ACE  may  be  found ;  and  if  from  it,  and  from 
ABE,  its  supplement  (BE  being  supposed  to  be  joined)  BCE 
be  taken,  the  remainders  are  the  angles  at  the  base. 

Analysis.  Let  ABC  be  the  required  triangle,  and  let  AD, 
the  line  bisecting  the  vertical  angle,  be  produced  to  meet  the 
circumference  of  the  circumscribed  circle  in  E;  join  also  EC. 
Then  (III.  19)  the  circumscribed  circle  is  given,  since  the  base 
and  vertical  angle  are  given  ;  and  the  arc  BEC  is  given,  as  are 
also  its  half  EC,  and  the  chord  EC.     Now  the  triangles  AEO, 


EXERCISES.  239 

DEC  are  equiangular ;  for  the  angle  CEA  is  common,  and  (IH. 
18,  cor.  1)  BCE  is  equal  to  BAE  or  EAC.  Hence  AE  :  EC  : : 
EC  :  ED,  and  theiefore  AE.ED=ECl  Hence  (IH.  21)  it  is 
evident  that  if  EC  be  made  a  tangent  to  a  circle,  and  if 
through  the  extremity  of  the  tangent  a  line  be  drawn  cutting 
the  circle,  so  that  the  part  within  the  circle  may  be  equal  to 
AD,  DE  will  be  equal  to  the  external  part ;  whence  the  con- 
struction is  manifest. 

Prop.  XIX. — Proh. —  Given  the  straight  liius  drawn  from 
the  three  angles  of  a  trian;^  le  to  the  points  of  bisection  of  the 
opposite  sides  /  to  construct  the  triangle. 

Trisect  the  three  given  lines,  and  describe  the  triangle  ABC 
having  its  three  sides  respectively  equal  to  two  thirds  of  the 
three  given  lines ;  complete  the  parallelogram  ABEC,  and 
draw  the  diagonal  AE ;  produce  also 
CB,  making  BE  equal  to  BC ;  and 
join  FA,  FE;  AFE  is  the  required 
triangle. 

Produce  AB,  EB  to  G,  H.  Then 
(H.  14,  cor.)  AE,  BC  bisect  each  other 
in  D,  and  therefore  FD  is  equal  to 
one  of  the  given  lines,  for  BD  is  one 

third  of  it,  and  FB  two  thirds.  Again :  because  FB,  BC  are 
equal,  and  HB  pai-allel  to  AC,  FA  is  bisected  in  H,  and  HB  is 
half  of  AC  or  BE.  Hence,  HE  is  equal  to  another  of  the  given 
lines,  and  it  bisects  FA.  In  the  same  manner  it  would  be 
proved,  that  AG  is  equal  to  the  remaining  line,  and  that  it 
bisects  FE.     Hence  FAE  is  the  triangle  required. 

Method  of  Computation.  BD,  which  is  a  third  of  one  of 
the  given  lines,  bitects  AE,  a  side  of  the  triangle  ABE,  in 
•which  the  sides  AB,  BE  are  respectively  two  thirds  of  the  two 
remaining  lines.  Then  (I.  24,  cor.  6,  and  II.  12,  cor.)  2AD^= 
AB'+BE*— 2BD';  whence  AD,  and  consequently  AE  may  be 
found ;  and  in  the  same  manner  the  other  sides  may  be  com- 
puted. 

Prop.  XX.— Prob, —  Given  the  three  perpendiculars  of  a  tri- 
angle y  to  construct  it. 


240 


EXERCI8E8. 


A- 
B- 
C- 
D- 


E 


Let  A,  B,  C  be  three  given  straight  lines;  it  is  required  to 
describe  a  triangle  having  its  three  perpendiculars  respectively 
eqnJil  to  A,  B,  C. 

Take  any  straight  line  D,  and  describe  a  triangle  EFG.  hav- 
ing the  sides  FG,  P'E,  EG  third  i)ro- 
portionals  to  A  and  1),  B  and  D,  C 
and  D;  and  draw  the  perpendiculars 
EH,  GL,  FK. 

Then  the  rectangles  rG.A,EF.Bare 
equal,  each  being  equal  to  the  square 
ofD;  and  therefore  EF  :  FG  :  :  A  : 
B.  But  in  the  similar  triangles  EHF, 
GLF,  EF  :  FG  :  :  EH  :  GL;  where- 
fore EH  :  GL  : :  A  :  B ;  and  in  the 
same  manner  it  would  be  proved,  that 
EH  :  FK  :  :  A  :  C.  Hence  (IV.  7) 
if  EH  be  equal  to  A,  GL  is  equal  to  B,  and  FK  to  C ;  and 
EFG  is  the  triangle  requii-ed. 

But  if  EH  be  not  equal  to  A,  make  EM  equal  to  it,  and  draw 
NMO  parallel  to  FG,  and  meeting  EF,  EG,  produced,  if  neces- 
sary, in  N  and  O  ;  ENO  is  the  required  triangle.     Draw  OP 
perpendicular  to  EN.     Then  EM  :  EH  :  :  EO  :  EG,  and  OP 
GL  :  :  EO  :  EG ;  whence  (IV.  7  and  alternately)  EM  :  OP  : 
EH  :  GL ;  or,  by   the   foregoing   part,  EM  :  OP  :  :  A  :  B 
wherefore  (IV.  7)  since  EM  is  equal  to  A,  OP  is  equal  to  B 
and  it  would  be  proved  in  a  similar  manner,  that  the  perpen- 
dicular from  N  to  EO  is  equal  to  C. 

Method  of  Computation.  By  dividing  any  assumed  num- 
ber successively  by  A,  B,  C,  we  find  the  sides  of  the  triangle 
EFG,  and  thence  its  angles,  or  those  of  ENO;  whence,  since 
the  perpendicular  EM  is  given,  the  sides  are  easily  found. 

Or,  when  the  sides  of  EFG  are  found,  its  perpendicular  EH 
may  be  comj)Uted  in  the  manner  pointed  out  in  the  note  to  tlie 
twelfth  proposition  of  the  second  book.  Then  EH  :  A  : :  FG: 
NO  : :  EF  :  EN  :  :  EG  :  EO. 


Prop.  XXI. — Prob, —  Given  the  sum  of  the  legs  of  a  right- 
angled  triangle,  and  the  sum  of  the  hypothenuse,  and  the  per- 
pendicular to  it  from  the  right  angle;  to  construct  the  triangle. 


EXERCISES.  241 

Let  tho  sum  of  the  legs  of  a  right-angled  triangle  be  equal 
to  the  straight  line  A,  and  the  sum  of  the  hypothenuse  and 
perpendicular  equal  to  BC ;  it  is  required  to  construct  the  tri- 
angle. 

Find  (I.  24,  cor.  4)  a  straight  line  the  square  of  which  is  equal 
to  the  excess  of  the  square  of  BC 

above  that  of  A,  and  cut  off  BD      ~ 

equal  to  that  line ;  on  DC  as  diam- 
eter describe  a  semicircle,  and  draw 
EF  parallel  to  DC  ate  a  distance 
equal  to  BD  ;  join  either  point  of 
intersection,  E,  with  D  and  C ; 
DEC  is  the  required  triangle. 

Draw  the  perpendicular  EG,  which  (const.)  is  equal  to  BD. 
Then  (II.  4)  BC^  or  AHEG^  =  (DC-fEG)^=DC'^  +  EG=+ 
2DC.EG;  whence  A^=DC^  +  2DC.EG.  Also  (DE  +  EC)^= 
DE^  +  EC^+2DE.EC=DC^-|-2DC.EG,  because  (III.  11,  and  I. 
24,  cor.  1)  DC'=DE^+EC^  and  DC.GE^DE.EC,  each  being 
equal  to  twice  the  area  of  the  triangle  DEC.  Hence  (DE4- 
EC)-  =  A-;  wherefore  (I.  23,  cor.  3)bE  +  EC=A;  and  DEC 
is  the  triangle  required. 

Method  of  Computatiofi.  From  the  construction,  we  have 
BD  or  EG=  V(BC'—A').  Then  DC  =  BC-BD;  by  halving 
which  we  get  the  radius  of  the  semicircle ;  and  if  from  the 
square  of  the  radius  drawn  from  E,  the  square  of  EG  be  taken, 
and  if  the  square  root  of  the  remainder  be  successively  taken 
from  the  radius,  and  added  to  it,  the  results  will  be  the  seg- 
ments DG,  GC ;  from  which,  and  from  EG,  the  sides  (I.  24,  cor. 
1)  are  readily  computed. 

Prop.  XXII. — Prob. —  Given  the  base  of  a  triangle,  the  per- 
pendicular, and  the  difference  of  the  sides  ;  to  construct  it. 

Make  AB  equal  to  the  given  base,  and  parallel  to  it  draw 
CD,  at  a  distance  equal  to  the  given  perpendicular;  dravvBDP 
perpendicular  to  CD,  and  make  DF  equal  to  DB  ;  from  A  as 
center,  with  a  radius  AE  equal  to  the  given  difference,  describe 
the  circle  ELN  ;  through  B,  F  describe  any  circle  cutting  ELN" 
in  L,  N,  and  let  G  be  the  point  in  which  a  straight  line  drawn 
through  L,  N  cuts  FB  produced  ;  draw  the  tangent  GK,  and 
16 


242 


EXEECISES. 


draw  AKM  cutting   CD  in  M;  join   BM;  and   it  k  ^ident 

from  the  second  corollary  to  tbe 
ninth  proposition  of  the  third  book, 
that  AMB  is  the  required  triangle* 
Method  of  Computation.  From 
M  as  center,  with  MK  as  radius, 
describe  a  circle,  and  by  the  corol- 
lary referred  to, it  will  pass  through 
B  and  F,  Join  AG  and  produce 
BA  to  H.  Then  the  rectangle 
FG.GB=AG^-AK^  each  being 
equal  to  the  square  of  GK  ;  that  is, 
FB.BG+BG==AB^-fBG^— AK^ 
Take  awayBG^;  the:i  FB.BG=AB^-AK==(AB+AK)  (AB 
—  AK)=HB.EB.  Hence  BG  becomes  known.  Then,  in  the 
two  right-angled  triangles  GBA,  GKA,  the  angles  at  A  can  be 
computed,  and  their  difference  is  the  angle  MAB  in  the  re- 
quired triangle.  The  rest  is  easy,  if  the  perpendicular  from 
M  to  AB  be  drawn. 


Prop.  XXIII. — Prob. —  Given  the  base,  the  area,  and  the 
•ratio  of  the  sides  of  a  triangle;  to  construct  it. 

Let  AB  be  the  given  base,  and  (V.  3,  and  scho.)  find  the 
points  C,  D,  so  that  AC,  CB,  and  AD,  DB  are  in  the  ratio  of 

the  sides ;  on  CD  as  diame- 
ter describe  the  semicircle 
CED,  and  (II.  5,  scho.)  to 
AB  apply  a  parallelogram 
BF  double  of  the  given  area ; 
let  FG,  the  side  of  this  oppo- 
site to  AB,  produced  if  neces- 
sary, cut  the  semicJircle  in  E,  and  join  EA,  EB  j  EAB  is  the  re- 
quired triangle. 

For  (Ex.  6,  cor.)  AE  is  to  EB  as  AC  to  CB ;  that  is,  in  the 
given  ratio.  Also  (1. 15,  cor.  4)  AEB  is  half  of  the  parallelo- 
gram BF,  which  is  double  of  the  given  area.  Therefore  AEB 
is  on  the  given  base,  is  of  the  given  area,  and  has  its  sides  in 
the  given  ratio. 

Scho.  If  FG,  or  FG  produced,  do  not  meet  the  semicircle, 


EXERCISE8.  243 

the  problem  is  impossible ;  if  it  cut  it  in  E  and  E^,  there  will 
be  two  triangles  essentially  different,  each  of  which  will  answer 
the  conditions  of  the  problem  ;  if  it  touch  the  semicircle,  there 
will  be  only  one  triangle,  and  it  will  be  the  greatest  possible 
with  the  base  of  the  given  magnitude,  and  the  sides  in  the 
given  ratio ;  and  hence  we  have  the  means  of  solving  the 
problem  in  which  it  is  required  to  construct  a  triangle  on  a 
given  base,  having  its  sides  in  a  given  ratio,  and  its  area  a 
maximu7n. 

Method  of  Computation.  Join  the  center  H  with  E,  and 
draw  the  perpendicular  EK.  Then,  let  m  '.  n  :  '.  AC  :  CB, 
and  consequently  m,  :  n  :  :  AD  :  DB ;  then,  from  these,  by 
composition  and  division,  we  get  m+n  :  7i  :  :  AB  :  CB,  and 
m—n  '.  n  :  :  AB  :  DB;  whence  DC  and  its  half,  the  radius 
of  the  circle,  become  known.  EK  also  is  found  by  dividing 
double  of  the  area  by  AB.  Then,  in  the  triangle  EKH,  KH 
can  be  found,  and  thence  AK  and  KB ;  and,  by  means  of  them 
and  EK,  the  sides  EA,  EB  may  be  computed.  If  E''  be  taken 
as  the  vertex,  the  method  of  computation  is  almost  the  same, 
and  is  equally  easy. 

Prob.  XXIV. — Prob. —  Given  the  base  of  a  triangle,  the  ver- 
tical angle,  and  the  rectangle  of  the  sides  ;  to  construct  it. 

On  the  given  base  describe  a  segment  containing  an  angle 
€qual  to  the  given  angle;  to  the  diameter  of  the  circle  of  which 
this  segment  is  a  part,  and  to  the  lines  containing  the  given 
rectangle,  find  a  fourth  proportional;  this  proportional  (Ex.  3) 
is  the  perpendicular  of  the  triangle  ;  and  the  rest  of  the  solu- 
tion is  effected  by  means  of  the  fifteenth  proposition  of  these 
Exercises. 

Prop.  XXV. — Prob, — To  divide  a  given  triangle  into  two 
parts  in  a  given  ratio,  by  a  straight  line  parallel  to  one  of  the 
sides. 

Let  ABC  be  a  given  triangle ;  it  is  required  to  divide  it  into 
two  parts  in  the  ratio  of  the  two  straight  lines,  m,  n,  by  a 
straight  line  parallel  to  the  side  BC. 

Divide  (V.  3,  scho.)  AB  in  G,  so  that  BG  :  GA  :  :  m  :  w, 
and  between  AB,  AG  find  (V.  11)  the  mean  proportional  AH' 


244 


EXERCISES. 


draw  HK  parallel  to  BC ;  ABC  is  divided  by  HK  in  the  man- 
ner required. 

For  (V.  14,  scho.)  since  the  three 
straight  lines  AB,  All,  AG  are  propor- 
tionals, AB  is  to  AG  as  the  triangle  ABC 
to  AHK ;  whence,  by  division,  BG  is  to 
AG,  or  (const.)  m  is  to  w,  as  the  quadrilat- 
eral BCKH  to  the  triangle  AHK. 

In   practice,  the   construction  is  easily 

and  elegantly  effected,  when  the  triangle 

is  to  be  divided  either  into  two  or  more 

"^  parts  proportional  to  given  lines,  by  divid- 

n  ing  AB  into  parts  proportional  to  those 

lines,  and  through   the  points  of  section 

drawing  perpendiculars  to  AB,  cutting  the  arc  of  a  semicircle 

described  on  AB  as  diameter;  then  by  taking  lines  on  AB, 

terminated  at  A,  and  severally  equal  to  the  chords  drawn  from 

A  to  the  points  of  section  of  the  arc,  the  points  on  AB  wall  be 

obtained  through  which  the  parallels  to  BC  are  to  be  drawn. 

The  reason  is  evident  from  the  foregoing  proof  in  connection 

with  the  corollary  to  the  eighth  proposition  of  the  fifth  book. 

Cor.  Hence  a  given  triangle  can  be  divided  into  two  parts 
in  a  given  ratio,  by  a  straight  line  parallel  to  a  given  straight 
line. 

Also,  a  triangle  can  be  divided  into  two  parts  in  a  given  ra- 
tio, by  a  straight  line  drawn  through  a  given  point  in  one  of 
the  sides;  and  a  given  quadrilateral  can  be  divided  into  tw^o 
parts  in  a  given  ratio,  by  a  straight  line  jsarallel  to  one  of  its 
sides. 


Prop.  XXVI. — Peob. — From  a  given  point  in  one  of  the 
sides  of  a  given  triangle,  to  draw  two  straight  lines  trisecting 
the  triangle. 

.  Let  ABC  be  the  given  triangle,  and 

D  the  given  point ;  then,  if  BD  be  not 
less  than  one  third  of  BC,  to  BD,  to  a 
third  part  of  BC,  and  to  the  perpendic- 
ular from  A  to  BC,  find  a  fourth  pro- 
portional, and  at  a  distance  equal  to  it 


EXERCISES.  245 

draw  a  parallel  to  BC,  cutting  BA  in  E,  and  join  ED  ;  the  tri- 
angle BED  is  evidently  (V.  10,  cor.,  and  I.  15,  cor.  4)  a  third 
part  of  ABC  ;  and  CDF  will  be  constructed  in  a  similar  man- 
ner, if  CD  be  not  less  than  a  third  of  BC. 

If  DC,  one  of  the  segments,  be  less  than  a  third  of  BC,  the 
triangle  BDE  is  constructed  as  before,  but  the  rest  of  the  pre- 
ceding solution  fails,  as  the  second  parallel  would  fall  above 
the  triangle.  In  this  case,  cut  off  BA  between  E  and  A,  a  part 
equal  to  BE,  and  call  it  EG;  then,  if  DG  be  joined,  the  trian- 
gle ABC  is  trisected  by  DE,  DG. 

Scho,  It  is  easy  to  see  that  this  method  may  be  readily  ex- 
tended to  the  division  of  a  triangle  into  more  equal  parts  than 
three,  or  into  parts  proportional  to  given  magnitudes,  by 
straight  lines  drawn  from  a  given  point  in  one  of  the  sides. 

Prop.  XXVII. — Theor. — If  the  sides  of  a  right-angled  tri- 
angle  be  continual  proportionals,  the  hypothenure  is  divided  in 
extreme  and  mean  ratio  by  the  perpendicular  to  it  from  the 
right  angle  /  and  the  greater  segment  is  equal  to  the  less  or  re- 
mote side  of  the  triangle. 

Let  ABC  be  a  triangle  right-angled  at  A,  and  let  AD  be  per- 
pendicular to  BC  ;  then  if  CB  :  BA 
:  :  BA  :  AC,  BC  is  divided  in  ex-  ■*■ 

treme  and  mean  ratio  in  D,  and  BD 
is  equal  to  AC. 

For  (hyp.)  CB  :  BA  : :  BA  :  AC, 
and  (V.  8,  cor.)  CB  :  BA  : :  BA  : 
BD ;    therefore   BA  :  AC  :  :  BA  : 

BD,  and  AC  is  equal  to  BD.  Again  (V.  8,  cor.),  BC  :  CA  :  : 
CA  :  CD,  or  BC  :  BD  :  :  BD  :  DC,  and  therefore  (V.  def  3) 
BC  is  divided  in  extreme  and  mean  ratio  in  D. 

Scho.  Conversely,  if  BC  :  BD  : :  BD  :  CD,  and  if  BAC  be  a 
right  angle,  and  DA  perpendicular  to  BC;  CB  :  BA  : :  BA  : 
AC,  and  BD  is  equal  to  AC.  For  (hyp.)  CB  :  BD  : :  BD  : 
CD,  and  (V.  8,  cor.)  CB  :  CA  : :  CA  :  CD ;  wherefore  BD^  is 
equal  to  CA',  each  (V.  9,  cor.)  being  equal  to  the  rectangle 
BC.CD,  and  therefore  BD  is  equal  to  CA.  Again  (V.  8),  CB  : 
BA  :  :  BA  :  BD  or  AC. 


246  EXERCISES. 

Prop.  XXVllL — Prob. —  Given  the  angles  and  diagonals 
of  a  parallelogram  ;  to  construct  it. 

On  one  of  the  diagonals  describe  a  segment  of  a  circle  con- 
taining  an  angle  equal  to  the  given  angle  at  either  extremity  of 
the  other ;  from  the  middle  point  of  this  diagonal  as  center, 
with  half  the  other  diagonal  as  radius,  describe  an  arc  cutting 
the  arc  of  the  segment ;  through  the  extremities  of  the  first 
diagonal  draw  four  straight  lines,  two  to  the  intersection  of  the 
arcs,  and  two  parallel  to  these ;  the  parallelogram  thus  formed 
is  easily  proved  to  be  the  one  required. 

Prop.  XXIX. — Prob. —  Given  the  vertical  angle  of  a  trian- 
gle^  and  the  radii  of  the  circles  inscribed  in  the  parts  into  which 
the  triangle  is  divided  by  the  perpendicular  y  to  construct  the 
triangle. 

Take  any  straight  line  ABC,  and  through  any  point  B  draw 
the  perpendicular  BD  ;  make  BA,  BC  equal  to  the  given  radii, 
and  let  E,  F  be  the  angular  points,  remote  from  B,  of  squares 

described  on  AB,  BC  ;  join  EF,  and 
on  it  describe  the  segment  EDF, 
containing  an  angle  equal  to  half  the 
given  vertical  angle ;  let  the  perpen- 
dicular cut  the  arc  EDF  in  D,  and 
join  DE,  DF  ;  draw  DG,  DH  making 
the  angles  EDG,  FDH  respectively 
equal  to  EDB,  FDB;  DGH  is  the 
required  triangle. 

For  (I.  14)  perpendiculars  drawn 
from  E  to  DB,  DG  are  equal,  and 
each  of  them  is  equal  (const.)  to  the  perpendicular  from  E  to 
GB.  Each  of  them  therefore  is  equal  to  the  given  radius  AB 
and  a  circle  described  from  E  at  the  distance  of  one  of  these  is 
inscribed  in  the  triangle  DGB.  In  the  same  manner  it  would 
be  shown,  that  a  circle  described  from  F  as  center,  with  the 
other  given  radius,  would  be  inscribed  in  DBII.  Hence,  since 
the  angle  GDH  is  double  of  EDF,  GDII  is  equal  to  the  given 
vertical  angle,  and  the  triangle  GDH  answers  the  conditions 
of  the  question. 

Scho.  The  preceding  solution  is  strictly  in  accordance  with 


EXEECI8E8.  24:T 

the  enunciation,  taken  in  its  limited  sense.  There  will  be  in- 
teresting variations,  however,  if  we  regard  the  given  circles, 
not  merely  as  inscribed.,  but  as  those  which  touch  all  the  sides 
of  each  of  the  right-angled  triangles,  either  internally  or  exter- 
nally. These  variations  will  be  obtained  by  giving  the  squares 
on  the  radii  every  possible  variety  of  position  in  the  four  right 
angles  formed  by  the  intersection  of  AC,  DB ;  and  the  solution 
will  obtain  complete  generality  by  taking  into  consideration 
both  the  points  in  which  BD  cuts  the  circle  of  which  EF  is  a 
chord. 

Prop.  XXX. — Theo^, —  The  area  of  a  triangle  ABC  ts 
equal  to  half  the  continued  product  of  two  of  its  sides,  AB, 
BC,  a?id  the  sine  of  their  contained  angle  B,  to  the  radius  I. 

Draw  the  perpendicular  AD.     Then  (Trig.  1,  cor.)  AD  = 
AB  X  sin  B.     Multiply  by  BC,  and  take 
Liiif  the  product ;  and  (I.  23,  cor.  6)  we 
have  the  area  equal  to  ^  AB  x  BC  X  sin  B. 

Cor.  Hence  (I.  15,  cor.  1)  the  area  of  a 
parallelogram  is  equal  to  the  continual 
product  of  two  contiguous  sides,  and  the 
sine  of  the  contained  angle. 

jScho.  From  this  proposition,  and  from  the  third  and  sixth 

corollaries  to  the  twenty-fifth  proposition  of  the  third  book,  we 

can  derive  neat  algebraic  expressions  for  the  radii  of  the  four 

circles,  each  touching  the  three  sides  of  a  triangle.     Thus,  by 

dividing  the  expression  for  the  area  by  s,  we  find,  according  to 

the  third  corollary,  that  the  radius  of  the  inscribed  circle  is 

/(s — a)  (s—b)  (s—c)        T    ...  ,      T   'T 

equal  to  ,{/ —^ —-^ ~ .      In  like  manner,  by  dividing 

the  expression  for  the  area  successively  by  s — a,  s—b,  s — c,  we 
find,  according  to  the  sixth  corollary,  that  the  radii  of  the  cir- 
cles touching  a,  b,  c,  externally,  are  respectively, 

/s{s-b){s-c)/s{s-a){s_-c)^  and  ^ ^  (s-a)  (s-b) ^ 
r  s—a  y  s—b  r  s  —  o 

By  taking  the  continual  product  of  these  four  expressions, 
and  contracting  the  result,  we  get  s  {s — a)  (s — b)  (s — c),  which. 
is  equal  to  the  square  of  the  area ;  and  hence,  by  expressing 


248  EXERCISES. 

this  inwards,  we  have  the  following  remarkable  theorem  :  The 
continual  product  of  the  radii  of  the  four  circles^  each  of  which 
touches  the  three  sides  of  a  triangle,  or  their  prolongations,  is 
equal  to  the  second  power  of  the  area,  which  is  proven  by  the 
next  proposition. 

Prop.  XXXI. — Theor. — Let  a,  b,  c  he  the  sides  of  a  trian- 
gle, and  s  half  their  sum  ;  the  area  is  equal  to  the  square  root 
of  the  continual  product  ofs,  s— a,  s — b,  and  s — c. 

It  was  proved  in  the  second  corollary  to  the  ninth  proposi- 
tion, Plane  Trigonometry,  that  the  sine  of  twice  any  angle  is 
twice  the  product  of  the  sine  and  cosine  of  the  angle.  Hence, 
by  multiplying  together  the  values  of  sin-^A  and  cos^A, 
given  in  the  corollaries  to  the  sixth  and  seventh  propositions, 
Plane  Trigonometry,  and  doubling  the  result,  we  get  sin  A= 

2  \/\s  (s-a)  (s-b)  (s-c)]      ^^        ,       , 

-j^ -.  Now,  by  the  precedmg  proposi- 
tion, the  area  of  a  triangle  is  found  by  multiplying  the  sine  of 
one  of  its  angles  by  the  sides  containing  it,  and  taking  half  of 
the  product;  multiplying,  therefore,  the  value  now  found  for 
sin  A,  by  be,  and  taking  half  the  product,  we  find  the  area  to 
be  V[s  {s — a)  {s — b)  (s — c)].  This  proposition  is  much  used  in 
surveying  coasts  and  harbors. 

Prop.  XXXTI. — Prob, — A  semicircle  ACB  being  given,  and 
other  semicircles  being  described  as  in  the  diagram,  /  it  is  re- 
quired  to  find  the  sum  of  the  areas  of  all  those  inscribed  semi- 
circles. 

Cii'cles   (V.   14),  and  consequently  semicircles,  are   as  the 

squares  of  their  diameters  or  of 
their  radii.  Now  the  square  of 
GD  is  half  the  square  of  DF  or 
CF,  and  therefore  the  semicircle 
DFE  is  half  of  ACB.  For  the 
same  reason  HGK  is  half  of  DFE; 
and  universally,  each  semicircle  is 
half  of  the  one  in  which  it  is  in- 
scribed. Hence  the  entire  amount  will  be  the  sum  of  the 
infinite  series  iACB-f-J  ACB+i  ACB+^  ACB+etc. ;    and 


EXERCISES. 


249 


therefore  (IV.  19)  i  ACB-i  ACB  :  ^ACB  ::  iACB  :  ACB, 
llie  required  sum ;  and  it  thus  appears  that  the  sum  of  all  the 
inscribed  semicircles  is  equivalent  to  the  given  semicircle. 

Pkop.  XXXIII— Theor.— J;z  any  triangle,  the  center  of  the 
circumscribed  circle,  the  point  in  which  the  three  perpendiculars 
intersect  one  another,  and  the  point  of  intersection  of  the 
straight  lines  drawn  from  the  angles  to  bisect  the  opposite  sides, 
lie  all  in  the  same  straight  line. 

Let  ABC  be  a  triangle,  and  let  the  two  perpendiculars  AD, 
CE  intersect  in  F ;  bisect  AB,  BC  in  H,  G,  and  draw  AG,  CH 
intersecting  in  K  ;  draw  also  GI,  HI  perpendicular  to  BC,  BA, 
and  intersecting  in  I.  Then  (Ex. 
7)  F  is  the  intersection  of  the  three 
perpendiculars,  K  (III.  1,  cor.  2) 
the  intersection  of  the  three  lines 
drawn  from  the  angles  to  bisect 
the  opposite  sides,  and  (III.  25, 
cor.  2)  I  is  the  center  of  the  cir- 
cumscribed circle.  Join  FK,  KI ; 
FKI  is  a  straight  line. 

Join  HG;  it  is  (V.  2  and  3) 
parallel  to  AC,  and  is  half  of  it.  Also  the  triangles  ACF, 
GHI  are  (I.  16,  cor.  3)  equiangular,  and  therefore  GI  is  half  of 
AF.  So  likewise  (III.  1,  scho.)  is  KG  of  KA.  Hence  the  two 
triangles  AKF,  GKI  have  the  alternate  angles  KAF,  KGI 
equal,  and  the  sides  about  them  proportional ;  therefoi-e  (V.  6) 
the  angles  AKF,  GKI  are  equal,  and  (1. 10,  cor.)  since  AKG  is 
a  straight  line,  FKI  is  also  a  straight  line. 

Scho.  It  is  plain  (V.  3)  that  FK  is  double  of  KI.  We  have 
also  seen  that  AF  is  double  of  GI.  Hence  it  appears,  that  the 
distance  between  any  of  the  angles  and  the  point  of  intersec- 
tion of  the  three  perpendiculars  is  double  of  the  perpendicular 
drawn  from  the  center  of  the  circumscribed  circle  to  the  side 
opposite  to  that  angle. 

Prop.  XXXIV. — Theor. — Straight  lines  drawn  from  the 
angles  of  a  triangle  to  the  points  in  which  the  opposite  sides 
touch  the  inscribed  circle,  all  pass  through  the  same  point. 


250 


EXERCISES. 


Let  ABC  be  a  triangle,  and  D,  E  the  points  in  which  the 
sides  AB,  AC  touch  the  inscribed  circle ;  draw  BFE,  CFD ; 
draw  also  AFG  cutting  BC  in  G ;  G  is  the  point  in  which  BC 
touches  the  inscribed  circle. 

If  possible,  let  another  point  K  be  the  point  of  contact,  and 

draw  DH,DI  parallel  to  BC, 
CA.  Then  in  the  similar  tri- 
angles FDI,  FCE,  FD  :  DI :  : 
FC  :  CE,  or  CK  ;  and  in  the 
similar  triangles  FDH,  FGC, 
DH  :  DF  : :  GC  :  FC  ;  from 
which  and  from  the  preced- 
ing analogy  we  get,  ex  cequo^ 
DH  :  DI : :  CG  :  CK.  Again, 
BD  :  DI  : :  BA  :  AE  or  AD 
: :  BG  :  DH.  Hence,  altern- 
ately, and  by  inversion,  BG  :  BD  : :  DH  :  DI  :  whence  (IV.  7) 
BG  :  BD  or  BK  : :  CG  :  CK,  or  alternately,  BG  :  CG  : :  BK  : 
CK;  and  by  composition,  BC  :  CG  ::  BC  :  CK ;  and  there- 
fore CG,  CK  are  equal ;  that  is,  G  and  K  coincide,  and  AFG 
passes  through  the  point  in  which  BC  touches  the  circle. 

Prop.  XXXV. — Theor. — 171  a  triangle,  the  sum  of  the  per- 
pendiculars draion  from  the  center  of  the  circumscribed  circle 
to  the  three  sides  is  equal  to  the  sum  of  the  radii  of  the  in- 
scribed and  circumscribed  circles. 

Let  ABC  be  a  triangle,  having  its  sides  bisected  in  D,  E,  F, 
by  perpendiculars  meeting  in  G,  the  center  of  the  circum- 
scribed circle  ;  the  sura  of  GD, 
GE,  GF  is  equal  to  the  sum  of 
the  radii  of  the  inscribed  and 
circumscribed  circles. 

Join  GA,  GB,  GC,  and  DE, 
EF,  FD.  Then,  putting  a,  b,  c 
to  denote  the  sides  opposite  to 
the  angles  A,  B,  C,  we  have 
(V.  2  and  3)  FE^^a,  FD=i6, 
and  DE=ic;  and  (HI.  11) 
Bince  AEG,  AFG  are  right  angles,  a  circle  may  be  described 


EXERCISER. 


251 


about  the  quadrilateral  AEGF.  For  a  like  reason  circles  may- 
be described  about  BDGF  and  CDGE.  Hence  (Ex.  4)  FE. 
AG=AF.GE  +  AE.FG;  or,  by  doubling,  a.AG=c.GEi-b.FG. 
In  the  same  manner,  it  would  be  shown,  since  AG,  BG,  CG  are 
equal,  that  ^>.AG=c.GD  +  a.FG,  and  c.AG  =  a.GE  +  ^..GD. 
Hence,  by  addition,  (a-|-5+c)AG  =  (a+c)GE+(ff  +  5)GF  + 
(54-c)GD,  Now  5.GE  is  evidently  equal  to  twice  the  triangle 
AGO,  c.GF  equal  to  twice  AGE,  and  a.GD  equal  to  twice 
BGC ;  also,  denoting  the  radius  of  the  inscribed  circle  by  r, 
we  have  (IH.  25,  cor.  3),  r{a+b-\-c)  equal  to  twice  the  area  of 
the  triangle  ABC,  and  consequently  r{a-\-b-i-c)=b.GE  +  c.GF 
+  a.GD.  Hence,  by  addition,  {a+b-\-c)AG-\-r{aA-b-t-c)  = 
(a+5+c)GE+(a+5+c)GF+(a+6+c)GD;  and  consequently 
AG4-r=GE+GF+GD. 

Cor.  Since,  by  the  scholium  to  proposition  thirty-third  of 
these  Exercises,  the  parts  of  the  three  perpendiculars  of  the 
trianffle,  between  their  common  intersection  and  the  three  an- 
gles,  are  respectively  double  of  GD,  GE,  GF,  the  sum  of  those 
parts  of  the  perpendiculars  is  equal  to  the  sum  of  the  diameters 
of  the  inscribed  and  circumscribed  circles. 


Prop.  XXXYI. — Theor, — If  on  the  three  sides  of  any  tri- 
angle equilateral  triangles  be  described^  either  all  externally^  or 
all  internally,  straight  lines  joining  the  centers  of  the  circles 
inscribed  in  those  three  triangles  form  an  equilateral  triangle. 

On  the  three  sides  of  any  triangle  ABC,  let'^the  equilateral 
triangles  ABD,  BCF,  CAE  be  de- 
scribed externally,  and  find  G,  H, 
K,  the  centers  of  the  circles  de- 
scribed in  those  triangles ;  draw 
GH,  HK,  KG  ;  GHK  is  an  equilat- 
eral triangle. 

Join  GA,  GB,  HB,  HC,  HF,  KC, 
AF.  Then  the  angle  FBC  is  two 
thirds  of  a  right  angle,  and  the  an- 
gles GAB,  GBA,  FBH,  BFH  each 
one  third.  The  triangles  FBH, 
ABG  are  therefore  similar,  and  (V. 
3)  FB  :  BH  : :  BA  :  BG ;  whence,  alternately,  FB  :  BA  : : 


252  EXERCISES. 

HB  :  BG;  that  is,  in  the  triangles  FBA,  HBG  the  sides  about 
the  angles  FBA,  HBG  are  proportional;  and  these  angles  aj-e 
equal,  each  of  thorn  being  equal  to  the  sum  of  the  angle  ABC 
and  two  thirds  of  a  right  angle.  Hence  (V.  6)  these  triangles 
are  equiangular ;  and  therefore  (V.  3)  FB  or  BC  :  FA  : :  BH 
:  HG;  and  it  would  be  shown  in  the  same  manner,  by  means 
of  the  triangles  ACF,  KCH,  that  FC  or  BC  :  FA  : :  CH  : 
HK ;  therefore  (IV.  1)  BH  :  HG  : :  CH  :  HK.  But  BH  is 
equal  to  CH,  and  therefore  (IV.  6)  HG  is  equal  to  HK ;  and  it 
would  be  demonstrated  in  a  similar  manner,  that  HG,  HK  are 
each  equal  to  GK. 

If  the  equilateral  triangles  were  described  on  the  other  sides 
of  the  lines  AB,  BC,  CA,  the  angles  ABF,  GBH  would  be  the 
difference  between  ABC  and  two  thirds  of  a  right  angle ;  but 
the  rest  of  the  proof  is  the  same. 

Sc/io.  If  ABC  exceed  aright  angle  and  a  third,  the  sum  of  it 
and  two  thirds  of  a  right  angle  is  greater  than  two  right  angles. 
In  that  case,  the  angles  ABF,  GBH,  understood  in  the  ordinary- 
sense,  are  each  the  difference  between  that  sum  and  four  right 
angles.  If  ABC  be  a  right  angle  and  a  third,  the  sum  becomes 
two  right  angles,  and  FB,  BA  are  in  the  same  straight  line,  as 
are  also  HB,  BG.  In  this  case  it  is  proved  as  before,  that  FB  : 
BA  : :  HB  :  BG ;  and  then  (IV.  11)  FB  or  BC  :  FA  ::  HB : 
HG.     The  rest  of  the  proof  would  proceed  as  above. 

It  may  be  remarked  that  if  an  equilateral  triangle  be  de- 
Bcribed  on  a  stj'aight  line,  and  if  on  the  two  parts  into  which  it 
is  divided  at  any  point,  other  equilateral  triangles  be  described, 
lying  in  the  opposite  direction,  the  lines  joining  the  centers  of 
the  three  equilateral  triangles  will  also  form  an  equilateral  tri- 
angle. The  connection  of  this  and  the  proposition  will  be  per- 
ceived by  supposing  two  angles  of  the  triangle  continually  to 
diminish,  till  they  vanish,  as  the  triangle  may  thus  be  conceived 
to  become  a  straight  line. 


TO    BE    PROVEN. 

1.  The  least  straight  line  that  can  be  drawn  to  another 
etraight  line  from  a  point  without  it,  is  the  perpendicular  to  it ; 
of  others,  that  which  is  nearer  to  the  perpendicular  is  less  than 


EXERCISES.  253 

one  more  remote ;  and  only  two  equal  straight  hnes  can  be 
drawn,  one  on  each  side  of  the  perpendicular. 

2.  Of  the  triangles  formed  by  drawing  straight  Hnes  from  a 
point  within  a  parallelogram  to  the  several  angles,  each  pair 
that  have  opposite  sides  of  the  parallelogram  as  bases,  are  half 
of  it. 

3.  If,  in  proceeding  round  an  equilateral  triangle,  a  square, 
or  any  regular  polygon,  in  the  same  direction,  points  be  taken 
on  the  sides,  or  the  sides  produced,  at  equal  distances  from  the 
several  angles,  a  similar  rectilineal  figure  will  be  formed  by 
joining  each  point  of  section  with  those  on  each  side  of  it. 

4.  If  the  three  sides  of  one  triangle  be  perpendicular  to  the 
three  sides  of  another,  each  to  each,  the  triangles  are  equi- 
angular. 

5.  A  trapezoid,  that  is,  a  trapezium  having  two  of  its  sides 
parallel,  is  equivalent  to  a  triangle  which  has  its  base  equal  to 
the  sum  of  the  parallel  sides,  and  its  altitude  equal  to  their  per- 
pendicular distance. 

6.  Given  the  segments  into  which  the  line  bisecting  the  ver- 
tical angle  divides  the  base,  and  the  difference  of  the  angles  at 
the  base  ;  to  construct  the  triangle,  and  compute  the  sides. 

7.  Within  or  without  a  triangle,  to  draw  a  straight  line  par- 
allel to  the  base,  such  that  it  may  be  equivalent  to  the  parts 
of  the  other  sides,  or  of  their  continuations,  between  it  and  the 
base. 

8.  Given  the  perpendicular  of  a  triangle,  the  diffei*ence  of  the 
segments  into  which  it  divides  the  base,and  the  difference  of 
the  angles  at  the  base ;  to  construct  the  triangle. 

9.  In  the  figure  for  the  first  corollary  to  the  twenty-fourth 
proposition  of  the  first  book,  prove  that  CD  is  perpendicular  to 
AH,  or  a  line  from  C  to  F  is  perpendicular  to  CD. 

10.  The  angle  made  by  two  chords  of  a  circle,  or  by  their 
continuations,  is  equal  to  an  angle  at  the  circumference  stand- 
ing on  an  arc  equivalent  to  the  sum  of  the  arcs  intercepted  be- 
tween the  chords,  if  the  point  of  intersection  be  within  the 
circle,  or  to  their  difference,  if  it  be  without ;  (2)  the  angle 
made  by  a  tangent  and  a  line  cutting  the  circle,  is  equal  to  aa 
angle  at  the  circumference  on  an  arc  equivalent  to  the  differ- 
ence of  the  arcs  intercepted  between  the  point  of  contact  and 


254  EXEKCIPHa, 

the  other  line;  and  (3)  the  angle  made  by  two  tangents  is 
equal  to  an  angle  at  the  circumference  standing  on  an  arc 
equivalent  to  the  difference  of  those  into  which  the  circumfer- 
ence is  divided  at  the  points  of  contact. 

Cor.  If  a  tangent  be  parallel  to  a  chord,  the  arcs  between 
the  point  of  contact  and  the  extremities  of  the  chord  are  equal. 

11.  Given  the  sum  of  the  perimeter  and  diagonal  of  a  square ; 
to  construct  it. 

12.  On  a  given  straight  line  describe  a  square,  and  on  the 
side  opposite  to  the  given  line  describe  equilateral  triangles 
lying  in  opposite  directions ;  circles  described  throiigh  the  ex- 
tremities of  the  given  line,  and  through  the  vertices  of  these 
triangles,  are  equal. 

13.  To  inscribe  an  equilateral  ti'iangle  in  a  given  square. 

14.  Given  the  angles  and  the  two  opposite  sides  of  a  trape- 
zium ;  to  construct  it. 

15.  In  a  given  circle  to  place  two  chords  of  given  lengths, 
and  inclined  at  a  given  angle. 

16.  In  the  figure  for  the  twenty-fifth  proposition  of  the  third 
book,  if  AM,  BM  be  joined,  the  angle  AMB  is  half  of  ACB. 

17.  If,  in  proceeding  in  the  same  direction  round  any  trian- 
gle, as  in  3,  points  be  taken  at  distances  from  the  several 
angles,  each  equal  to  a  third  of  the  side,  the  triangle  formed  by 
joining  the  points  of  section  is  one  third  of  the  entire  triangle. 

18.  To  describe  a  square  having  two  of  its  angular  points  on 
the  circumference  of  a  given  circle,  and  the  other  two  on  two 
given  straight  lines  drawn  through  the  center.  Show  that 
there  may  be  eight  such  squares. 

19.  Given  the  vertical  angle  of  a  triangle,  and  the  segments 
into  which  the  base  is  divided  at  the  point  of  contact  of  the  in- 
scribed circle  ;  to  describe  the  triangle,  and  compute  the  sides. 

20.  If  any  three  angles  of  an  equilateral  pentagon  be  equal, 
all  its  angles  are  equal. 

21.  Given  two  sides  of  a  triangle,  and  the  difference  of  the 
sefrments  into  which  the  third  side  is  divided  by  the  perpen- 
dicular from  the  opposite  angle ;  to  construct  the  triangle. 

22.  Given  the  vertical  angle  of  a  triangle,  the  line  bisecting 
it,  and  the  perpendicular;  to  construct  the  triangle. 

23.  From  a  given  center  to  describe  a  circle,  from  which  a 


EXERCISES.  265 

straight  line,  given  in  position,  will  cut  off  a  segment  contain- 
ing an  angle  equal  to  a  given  angle. 

24.  In  a  given  triangle  to  inscribe  a  semicircle  having  its 
center  in  one  of  the  sides. 

25.  Through  three  given  points  to  draw  three  parallels, 
two  of  which  may  be  equally  distant  from  the  one  between 
them. 

26.  Given  an  angle  of  a  triangle,  and  the  radii  of  the  circles 
touching  the  sides  of  the  triangles  into  which  the  straight  line 
bisecting  the  given  angle  divides  the  triangle ;  to  construct  it. 

27.  In  a  rhombus  to  inscribe  a  square. 

28.  Given  the  lengths  of  the  two  parallel  sides  of  a  trapezoid, 
and  the  lengths  of  the  other  sides  ;  to  construct  it. 

29.  Given  one  of  the  angles  at  the  base,  and  the  segments 
into  which  the  base  is  divided  at  the  point  of  contact  of  the  in- 
scribed circle  ;  to  describe  the  triangle. 

30.  To  draw  a  tangent  to  a  given  circle,  such  that  the  part 
of  it  intercepted  between  the  continuations  of  two  given  diam- 
eters may  be  equal  to  a  given  straight  line. 

31.  To  draw  a  tangent  to  a  given  circle,  such  that  the  part 
of  it  between  two  given  tangents  to  the  cii'cle  may  be  equal  to 
a  given  straight  line. 

32.  Given  the  vertical  angle  of  a  triangle,  the  difference  of 
the  sides,  and  the  difference  of  the  segments  into  which  the 
line  bisecting  the  vertical  angle  divides  the  base  j  to  construct 
the  triangle. 

33.  Given  any  three  of  the  circles  mentioned  in  the  fifth 
■corollary  to  the  twenty-fifth  proposition  of  the  third  book ;  to 
describe  the  triangle. 

34.  A  straight  line  and  a  point  being  given  in  position,  it  is 
required  to  draw  through  the  point  two  straight  lines  inclined 
at  a  given  angle,  and  inclosing  with  the  given  line  a  space  of 
given  magnitude. 

35.  From  two  given  straight  lines  to  cut  off  equal  parts,  each 
of  which  will  be  a  mean  proportional  between  the  remainders. 

36.  A  square  is  to  a  regular  octagon  described  on  one  of  its 
sides,  as  1  to  2  (1-f  V2). 

37.  In  a  given  triangle  to  inscribe  a  parallelogram  of  a  given 
area. 


256  EXEECI8E8. 

38.  In  a  given  circle  to  inscribe  a  parallelogram  of  a  given 
area. 

39.  Through  a  given  point  between  the  lines  forming  a 
given  angle,  to  draw  a  straight  line  cutting  off  the  least  possi- 
ble triangle. 

40.  To  divide  a  given  straight  line  into  two  parts,  such  that 
the  square  of  one  of  them  may  be  double  of  the  square  of  the 
other,  or  may  be  in  any  given  ratio  to  it. 

41.  To  produce  a  given  straight  line,  so  that  the  square  of 
the  whole  line  thus  produced  may  be  double  of  the  square  of 
the  part  added,  or  in  any  given  ratio  to  it. 

42.  Given  the  area  of  a  right-angled  triangle,  and  the  sura  of 
the  legs  ;  to  construct  it. 

43.  Given  the  area  and  the  difference  of  the  legs  of  a  right- 
angled  triangle ;  to  construct  it. 

44.  Given  one  leg  of  a  right-angled  triangle,  and  the  remote 
segment  of  the  hypothenuse,  made  by  a  perpendicular  from  the 
right  angle ;  to  construct  the  triangle. 

45.  Given  the  base  of  a  triangle,  the  vertical  angle,  and  the 
side  of  the  inscribed  square  standing  on  the  base ;  to  describe 
the  triangle. 

46.  Given  the  base  of  a  triangle,  and  the  radii  of  the 
inscribed  and  circumscribed  circles  ;  to  construct  the  tri- 
angle. 

47.  To  draw  a  chord  in  a  circle  which  will  be  equal  to  one 
of  the  segments  of  the  diameter  that  bisects  it. 

48.  In  a  given  circle  to  draw  a  chord  which  will  be  equal  to 
the  difference  of  the  parts  into  which  it  divides  the  diameter 
that  bisects  it. 

49.  If  two  sides  of  a  regular  octagon,  between  which  two 
others  lie,  be  produced  to  meet,  each  of  the  produced  parts  is 
equivalent  to  a  side  of  the  octagon,  together  with  the  diagonal 
of  a  pquare,  described  on  the  side. 

50.  The  perimeter  of  a  triangle  is  to  the  base  as  the  perpen- 
dicular to  the  radius  of  the  inscribed  circle. 

51.  From  a  given  point  without  a  given  circle  to  draw  a 
straight  line  cutting  the  circle,  so  that  the  external  and  inter- 
nal parts  may  be  in  a  given  ratio. 

52.  Each  of  the  complements  of  the  parallelograms,  about 


EXERCISES.  257 

the  diagonal  of  a  parallelogram,  is  a  mean  proportional  between 
those  parallelograms. 

53.  Given  the  ratio  of  two  straight  lines,  and  the  difference 
of  their  squares ;  to  find  them. 

54.  The  square  of  the  perimeter  of  a  right-angled  triangle  is 
equivalent  to  twice  the  rectangle  under  the  sura  of  the  hy- 
pothenuse  and  one  leg,  and  the  sum  of  the  hypothenuse  and  the 
other. 

55.  The  quadrilateral  formed  by  straight  lines  bisecting  each 
pair  of  adjacent  sides  of  a  quadrilateral  is  a  parallelogram, 
which  is  half  of  the  quadrilateral ;  and  straight  lines,  joining 
the  points  in  which  the  sides  of  that  parallelogram  are  cut  by 
the  diagonals  of  the  primitive  figure,  form  a  quadrilateral  simi- 
lar to  that  figure  and  equivalent  to  a  fourth  of  it. 

66.  From  three  given  points  as  centers,  and  not  in  the  same 
straight  line,  to  describe  three  circles  each  touching  the  other 
two.     Show  that  this  admits  of  four  solutions. 

57.  To  add  a  parallelogram  to  a  rhombus,  such  that  the 
M'hole  figure  may  be  a  parallelogram  similar  to  the  one  added. 

58.  A  straight  line  being  given  in  position,  and  a  circle  in 
magnitude  and  position,  it  is  required  to  describe  two  equal 
circles  touching  one  another,  and  each  touching  the  straight 
line  and  the  circle. 

59.  The  squares  of  the  diagonals  of  a  quadrilateral  are  to- 
gether double  of  the  squares  of  the  straight  lines  joining  the 
points  of  bisection  of  the  opposite  sides. 

60.  In  a  given  rhombus  to  inscribe  a  rectangle  having  its 
sides  in  a  given  ratio. 

61.  If,  throu2rh  the  vertex  and  the  extremities  of  the  base  of 
a  triangle,  two  circles  be  described  intersecting  one  another  in 
the  base  or  its  continuation,  their  diameters  are  proportional  to 
the  sides  of  the  triangle. 

62.  To  draw  a  straight  line  cutting  two  given  concentric 
circles,  so  that  the  parts  of  it  within  them  may  be  in  a  given 
ratio. 

63.  From  a  given  point,  within  a  given  circle,  or  without  it, 
to  draw  two  straight  lines  to  the  circumference,  perpendicular 
to  one  another,  and  in  a  given  ratio.  "When  will  this  be  im- 
possible ? 

It 


258  EXERCISES. 

64.  If  a  straight  line  be  divided  in  extreme  and  mean  ratio, 
the  squares  of  the  whole  and  the  less  part  are  together  equiva- 
lent to  thi-ee  times  the  square  of  the  greater, 

65.  If  a  straight  line  be  cut  in  extreme  and  mean  ratio,  and 
be  also  bisected,  the  square  of  the  intermediate  part,  and  three 
times  the  square  of  half  the  line  are  equivalent  to  twice  the 
square  of  the  greater  part. 

66.  If  the  hypothenuse  of  a  right-angled  triangle  be  given, 
the  side  of  the  greatest  inscribed  square,  standing  on  the  hy- 
pothenuse, is  one  third  of  the  hypothenuse. 

67.  To  divide  a  given  semicircle  into  two  parts  by  a  perpen- 
dicular to  the  diameter,  so  that  the  radii  of  the  circles  inscribed 
in  them  may  be  in  a  given  ratio. 

68.  To  draw  a  straight  line  parallel  to  the  base  of  a  triangle, 
making  a  segment  of  one  side  equal  to  the  remote  segment  of 
the  other. 

69.  In  the  figure  for  the  tenth  proposition  of  the  second  book, 
the  square  of  the  diameter  of  the  circle,  passing  through  the 
points  F,  H,  D,  is  six  times  the  square  of  the  straight  line  join- 
ing P^D. 

VO.  Given  the  base,  the  area,  and  the  sum  of  the  squares  of 
the  sides  of  a  triangle ;  to  construct  it. 

71.  If,  from  the  extremities  of  the  hypothenuse  of  a  right- 
angled  triangle  as  centers,  arcs  be  described  passing  through 
the  right  angle,  the  hypothenuse  is  divided  into  three  segments, 
such  that  the  square  of  the  middle  one  is  equivalent  to  twice 
the  rectangle  of  the  others. 

72.  On  a  given  hypothenuse  to  describe  a  right-angled  tri- 
angle, such  that  the  difference  between  one  leg  and  the  adja- 
cent segment  of  the  hypothenuse  made  by  a  perpendicular  from 
the  right  angle,  may  be  a  maximum, 

73.  On  a  given  hypothenuse  to  construct  a  right-angled  tri- 
angle, such  that  one  segment  of  the  hypothenuse  made  by  the 
perpendicular  from  the  right  angle,  may  be  equivalent  to  the 
Bum  of  the  perpendicular  and  the  other  segment. 

74.  The  square  of  DH  (see  figure  for  III.  24)  is  equivalent  to 
the  rectangle  CF.BG ;  and  if  the  circles  touch  one  another  ex- 
ternally, DH  is  a  mean  proportional  between  their  diameters. 
Also  the  square  of  DH  is  equivalent  to  the  rectangle  CG.BF. 


EXERCISES. 


259 


75.  On  a  given  straight  line  to  describe  an  isosceles  triangle, 
having  the  vertical  angle  treble  of  each  of  the  angles  at  the 
base. 

V6.  If  in  the  diameter  of  a  circle  and  its  continuation  two 
points  be  taken  on  the  opposite  sides  of  the  center,  such  that 
the  rectangle  under  their  distances  from  the  center  may  be 
equivalent  to  the  square  of  the  radius,  any  circle  whatever  de- 
scribed through  these  points  bisects  the  circumference  of  the 
otlier  circle. 

'77.  To  find  a  point  from  which,  if  straight  lines  be  drawn  to 
three  given  points,  they  will  be  proportional  to  three  given 
straight  lines. 

78.  Given  the  segments  into  which  the  base  of  a  tnangle  is 
divided  by  two  straight  lines  trisecting  the  vertical  angle ;  to 
construct  it. 

79.  If  one  diagonal  of  a  quadrilateral  inscribed  in  a  circle  be 
bisected  by  the  other,  the  square  of  the  latter  is  equivalent  to 
half  the  sum  of  the  squares  of  the  four  sides. 

80.  To  divide  a  straight  line  into-  two  parts,  such  that  the 
squares  of  the  whole  and  one  of  the  parts  may  be  double  of  the 
square  of  the  other  part. 

81.  Given  the  segments  into  which  the  base  of  a  triangle  is 
divided  by  the  straight  line  bisecting  the  vertical  angle,  to 
construct  the  triangle  so  that  its  angle  adjacent  to  the  greater 
segment  may  be  either  of  a  given  magnitude,  or  a  maximum, 
and  in  each  case  to  compute  the  remaining  sides  and  angles. 

82.  To  draw  a  straight  line  bisecting  a  given  parallelogram, 
so  that  if  it  be  produced  to  meet  the  sides  produced,  the  ex- 
ternal triangles  will  have  a  given  ratio  to  the  parallelogram. 

83.  Through  a  given  point  to  draw  a  straight  line,  which, 
if  continued,  would  pass  through  the  point  of  intersection  of 
two  given  inclined  sti'aight  lines,  without  producing  those  lines 
to  meet. 

84.  If,  from  any  point  in  the  circumference  of  the  circle  de- 
scribed about  an  equilateral  triangle,  chords  be  drawn  to  its 
three  angles,  the  sum  of  their  squares  is  equivalent  to  six  times 
the  square  of  the  radius  of  the  same  circle,  or  to  twice  the 
square  of  a  side  of  the  triangle. 

85.  Given  the  difference  of  the  angles  at  the  base  of  a  trian- 


260  EXERCISES. 

gle,  the  difference  of  the  segments  into  which  the  hase  is 
divided  by  the  perpendicular,  and  the  ratio  of  the  sides ;  to 
construct  the  triangle. 

86.  Given  the  base  and  vertical  angle  of  a  triangle,  to  con- 
struct it  so  that  the  line  bisecting  the  vertical  angle  may  be  a 
mean  proportional  between  the  segments  into  which  it  divides 
the  base. 

87.  Given  two  sides  of  a  triangle,  and  the  ratio  of  the  base 
and  the  line  bisecting  tlie  vertical  angle  j  to  construct  the  tri- 
angle. 

88.  If  the  vertical  angle  of  a  triangle  be  double  of  one  of  the 
angles  at  the  base,  the  rectangle  under  the  sides  is  equivalent 
to  the  rectangle  under  the  base,  and  the  line  bisecting  the  ver- 
tical angle. 

89.  If  a  straight  line  be  divided  into  parts,  which,  taken  in 
succession,  are  continual  proportionals,  and  if  circles  be  de- 
scribed on  the  several  parts  as  diameters,  a  straight  line  which 
touches  two  of  the  circles  on  the  same  side  of  the  straight  line 
joining  their  centers,  will  touch  all  the  others. 

90.  Given  the  segments  into  which  the  base  is  divided  by 
the  straight  line  bisecting  the  vertical  angle,  and  the  angles 
which  that  straight  line  makes  with  the  base ;  to  construct  the 
triangle. 

91.  To  divide  a  given  circle  into  two  segments,  such  that 
the  squares  inscribed  in  them  may  be  in  a  given  ratio. 

92.  Through  a  given  point  in  the  base  of  a  given  isosceles 
triangle,  or  its  continuation,  to  draw  a  straight  line  such  that 
the  lines  intercepted  on  the  equal  sides,  or  their  continuations 
between  that  line  and  the  extremities  of  the  base,  may  have 
one  of  the  equal  sides  as  a  mean  proportional  between  them. 

93.  Through  a  point  in  tlie  circumference  of  a  given  circle,  to 
draw  two  chords,  such  that  their  rectangle  may  be  equivalent 
to  a  given  space,  and  the  chord  joining  their  other  extremities 
equal  to  a  given  straight  line. 

94.  In  any  triangle  the  radius  of  the  circvmiscribed  circle  is 
to  the  radius  of  the  circle  which  is  the  locus  of  the  vertex, 
when  the  base  and  the  ratio  of  the  sides  are  given,  as  the  dif- 
ference of  the  squares  of  those  sides  is  to  four  times  the  area. 

95.  The  difference  of  the  sides  of  a  triangle  is  a  mean  pro- 


KXERCI8K8.  261 

portional  between  the  clifFerence  of  the  segments  into  which 
the  base  is  divided  by  the  perpendicular,  and  the  difference  of 
those  into  which  it  is  divided  by  the  line  bisecting  the  vertical 
angle. 

96.  Let  the  angles  of  a  pai*allelogram  which  has  unequal 
sides  be  bisected  by  straight  lines  cutting  the  diagonals,  and 
let  the  points  of  intersection  be  joined ;  the  figure  thus  formed 
is  a  parallelogram,  which  has  to  the  proposed  parallelogram  the 
duplicate  ratio  of  that  which  the  difference  of  the  unequal  sides 
of  the  latter  has  to  their  sum. 

97.  A  circle  and  a  point  being  given,  it  is  required  to  de- 
scribe a  triangle  similar  to  a  given  one,  having  its  vertex  at  the 
given  point,  and  its  base  a  chord  of  the  given  circle. 

98.  Given  the  three  points  in  which  the  sides  of  a  triangle 
are  cut  by  the  perpendiculars  from  the  opposite  angles ;  to  con- 
struct the  triangle. 

99.  Given  the  ansjles  of  a  triansrle,  and  the  leno;ths  of  three 
straight  lines  drawn  from  the  angular  points  to  meet  in  an- 
other point ;  to  construct  the  triangle. 

100.  Given  the  base  of  a  triangle,  and  the  ratio  of  its  sides; 
to  construct  it,  so  that  the  distance  of  its  vertex  from  a  given 
point  may  be  a  maximum  or  minimum. 

101.  To  divide  {^circle  into  two  segments,  such  that  the  sum 
of  the  squares  inscribed  in  them  may  be  equivalent  to  a  given 
space. 

102.  Through  a  given  point,  with  a  given  radius,  to  describe 
a  circle  bisecting  the  circumference  of  a  given  circle. 

103.  With  a  given  radius  to  describe  a  cn-cle  bisecting  the 
circumferences  of  two  given  circles. 

104.  In  a  right-angled  triangle,  the  rectangle  under  the 
radius  of  the  inscribed  circle,  and  the  radius  of  the  circle 
touching  the  hypothenuse  and  the  legs  produced,  is  equivalent 
to  the  area.  So,  likewise,  is  the  rectangle  under  the  circles 
touching  the  legs  externally,  and  the  continuations  of  the  other 
sides. 

105.  If  three  straight  lines  be  continual  proportionals,  the 
6um  of  the  extremes,  their  difference,  and  double  the  mean  will 
be  the  hypothenuse  and  legs  of  a  right-angled  triangle. 

106.  From  two  given  centers,  to  describe  circles  having  their 


263  EXERCISES. 

radii  in  a  given  ratio,  and  the  part  of  their  common  tangent, 
between  the  points  of  contact,  equal  to  a  given  straiglit  line. 

107.  A  straiglit  line  and  two  points  equally  distant  from  it, 
on  the  same  side,  being  given  in  position,  it  is  required  to  draw 
through  the  points  two  straight  lines  forming  with  the  given 
line  the  least  isosceles  triangle  possible,  on  the  side  on  which 
the  points  ai"e. 

108.  To  describe  a  circle  touching  a  diameter  of  a  given  cir- 
cle in  a  given  point,  and  having  its  circumference  bisected  by 
that  of  the  given  one. 

109.  If  an  angle  of  a  triangle  be  60^,  the  square  of  the  oppo- 
site side  is  less  than  the  squares  of  the  other  two  by  their  rect- 
angle ;  but  if  an  angle  be  1 20°,  the  square  of  the  opposite  side 
is  greater  than  the  squares  of  the  others  by  their  rectangle. 

110.  In  the  figure  on  page  251,  prove  that  the  three  straight 
lines  joining  AF,  BE,  CD  are  all  equal. 

111.  The  chord  of  120°  is  equal  to  the  tangent  of  60°. 

112.  The  sines  of  the  parts  into  which  the  vertical  angle  of  a 
triangle  is  divided  by  the  straight  line  bisecting  the  base,  are 
reciprocally  proportional  to  the  adjacent  sides.  Show  from 
this  how  a  given  angle  may  be  divided  iuto  two  parts,  having 
their  sines  in  a  given  ratio. 

113.  The  diameter  of  the  circle  described  about  any  triangle 
is  equivalent  to  the  product  of  any  side  and  the  cosecant  of  the 
oj^posite  angle. 

114.  In  any  triangle  ABC,  the  radius  of  the  inscribed  circle 

.     ,  sin  A  B  sin  A  C  .  sin  i  A  sin  ^^  C 

IS  equivalent   to   a. t—: >  to   o. r-f, —  >  or   to 

^  cos  Y  A  cos  ^- 1> 

sin  4-  A  sin  4  B  £     n      .     -i         i  *    /• 

c. r-^ — ;  or,  finally,  to  the  cube  root  oi 

cos  -J  O 

abc  sin  ^  A  sin  ^  B  sin  -J  C  tan  ^  A  tan  |  B  tan  ^  C. 

115.  Given  the  sum  of  the  tangents,  and  the  ratio  of  the  se- 
cants, of  two  angles  to  a  given  radius ;  to  determine  the  angles 
geometrically  and  by  computation. 

116.  Find  an  angle,  such  that  its  tangent  is  to  the  tangent 
of  its  double,  in  a  given  ratio  ;  suppose  that  of  2  to  5. 

117.  If  a  spherical  triangle  be  right-angled  at  C,  and  any  two 
of  the  other  parts  be  made  s -.ccessively  108°  42'  and  87°  33' 
19",  what  will  be  the  values  o   the  remaining  parts? 


ADDENDUM. 


263 


118.  Let  any  three  parts  of  an  oblique-angled  spherical  trian- 
gle be  successively  87°  45^  96°  57'  48'^,  and  106°  53'  13'';  what 
will  be  the  values  of  the  other  parts? 

119.  When  the  three  sides  of  a  spherical  triangle  be  respect- 
ively, 34°  39'  44",  78°  27'  49",  and  134°  15'  23",  what  will  be 
the  surface  of  the  triangle  when  the  radius  of  the  sphere  is  16  ? 
"What  will  be  the  base  of  the  triangular  pyramid  subtended  by 
those  sides,  and  what  will  be  the  surface  and.  solidity  of  the 
spherical  pyramid  with  its  apex  at  the  center  of  the  sphere  ? 

120.  If  the  foregoing  values  be  the  magnitudes  of  the  angles 
of  a  spherical  triangle  when  the  radius  of  the  sphere  is  16,  what 
will  be  the  base  of  the  triangular  pyramid  subtended  by  the 
sides  of  the  spherical  triangle  ?  What  will  be  the  surface  of 
the  spherical  triangle,  and  the  surface  and  solidity  of  the 
spherical  j^yramid  with  its  apex  at  the  center  of  the  sphere  ? 

121.  If  the  earth  be  regarded  a  sphere  with  7973.8798-f  miles 
for  its  diameter,  and  a  spherical  pentagon  be  measured  on  its 
surface  having  341.78  miles,  309.25  miles,  278.64  miles,  173.97 
miles,  and  97  miles  for  its  sides  ;  and  the  angles  contained  by 
those  sides  be  respectively  74°  34'  19",  107°  09'  51",  41°  0' 
n'",  85°  17'  09",  and  76°  41'  35",  what  will  be  the  surface  of 
the  spherical  pentagon  ?  and  what  will  be  the  solidity  of  the 
pyramid  having  the  spherical  pentagon  for  its  base  and  its  apex 
at  the  center  of  the  earth  ? 


ADDENDUM, 

Illustrating  the  Third  Proof  for  the  Second  Corollary 
TO  THE  Seventeenth  Proposition  of  the  tixTii  Book, 
Elements  of  Euclid  and  Legendre. 

Circles  are  to  one  another  as  the  squares  described  upoft 
their  diameters  (V.  14),  consequently  squares  are  to  one  an- 
other as  the  circles  described  (V.  14,  cor.  2)  upon  their  sides; 
that  is,  there  is  a  ratio  of  equality  between  the  circle  and 
squares  which  have  the  same  straight  line  for  their  respective 
diameter,  side,  and  diagonal  •  therefore  the  circle  has  the  same 


264  ADDENDUM. 

arithmetical  proportion  to  the  inscribed  square  having  the 
diameter  for  its  diagonal,  as  the  circumscribed  square  having 
the  same  diameter  for  its  side  has  to  the  circle.  If  10  be  the 
diameter  of  the  circle,  and  x  ^6  the  area  of  the  circle,  100  will 
be  the  area  of  the  circumscribed  square,  and  50  will  be  the  area 
of  the  inscribed  square — and  we  have  the  arithmetical  propor- 
tion, 

100,  X,  50; 

and  the  geometrical  proportion, 

100  :  Y  :  :  50  :  -5-. 

From  the  first  proportion,  we  derive 

100— x=x— 50. 
From  the  second  proportion,  we  have 


100-x  =  2  (50- I). 


.-.  x-50  =  2  (50-|x)  =  100-x; 
or,  2p(;=:150,  hence  p^  =  75. 
The  arithmetical  proportion  gives, 

2x=150  or  pf=75 
Substituting  this  value  for  -/^  in  the  above  proportions,  \,lpl 
get 

100  :  75  :  :  50  :  371; 
and  100—75  =  75—50; 
and  100— 75  =  2(50— 371); 
and  100,  75,  50; 

/.  s                    ^          150      ^ 
or,  2  (75)  =  150,  or  75= =75. 

Li 

Hence,  the  circle  is  the  arithmetical  mean  between  the 
squares,  circumscribed  and  inscribed  about  it ;  or  three  fourths 
of  the  circumscribed  square;  or  three  times  square  of  the  ra- 
dius.    (See  Exercises,  def  7,  scho.  2). 

Thus  we  have  the  area  of  the  circle  expressed  by  a  finite 
quantity  instead  of  the  irrational  quantity  (V.  25,  scho.  1), 
giving  the  approximate  area  only  of  the  circle. 


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